Binomial Theorem worksheet for Class 11 with 13 problems on expansion, coefficients, and terms.

Since all user options (ExplainAnswer, TeachTopic, GeneratePractice) are false, I will skip all explanatory sections and proceed directly to solving the problems step by step, then provide only the final answers as required.

There are 13 problems in total. I’ll solve each one carefully.

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Problem 1: Expand $(2x - \frac{1}{x})^5$

Use binomial theorem: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here, $a = 2x$, $b = -\frac{1}{x}$, $n = 5$

Terms:

- $k=0$: $\binom{5}{0}(2x)^5(-\frac{1}{x})^0 = 1 \cdot 32x^5 \cdot 1 = 32x^5$
- $k=1$: $\binom{5}{1}(2x)^4(-\frac{1}{x})^1 = 5 \cdot 16x^4 \cdot (-\frac{1}{x}) = -80x^3$
- $k=2$: $\binom{5}{2}(2x)^3(-\frac{1}{x})^2 = 10 \cdot 8x^3 \cdot \frac{1}{x^2} = 80x$
- $k=3$: $\binom{5}{3}(2x)^2(-\frac{1}{x})^3 = 10 \cdot 4x^2 \cdot (-\frac{1}{x^3}) = -40x^{-1}$
- $k=4$: $\binom{5}{4}(2x)^1(-\frac{1}{x})^4 = 5 \cdot 2x \cdot \frac{1}{x^4} = 10x^{-3}$
- $k=5$: $\binom{5}{5}(2x)^0(-\frac{1}{x})^5 = 1 \cdot 1 \cdot (-\frac{1}{x^5}) = -x^{-5}$

So expansion:
$32x^5 - 80x^3 + 80x - \frac{40}{x} + \frac{10}{x^3} - \frac{1}{x^5}$

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Problem 2: Expand $(\sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}})^6$

Let $u = \sqrt{\frac{x}{a}}$, $v = \sqrt{\frac{a}{x}}$, so $(u + v)^6$

Note: $uv = \sqrt{\frac{x}{a} \cdot \frac{a}{x}} = \sqrt{1} = 1$

Binomial expansion:

$\sum_{k=0}^{6} \binom{6}{k} u^{6-k} v^k = \sum_{k=0}^{6} \binom{6}{k} u^{6-k} v^k$

But since $v = 1/u$, then $u^{6-k} v^k = u^{6-k} u^{-k} = u^{6-2k}$

Alternatively, write directly:

Term k: $\binom{6}{k} \left(\sqrt{\frac{x}{a}}\right)^{6-k} \left(\sqrt{\frac{a}{x}}\right)^k = \binom{6}{k} \left(\frac{x}{a}\right)^{(6-k)/2} \left(\frac{a}{x}\right)^{k/2} = \binom{6}{k} \left(\frac{x}{a}\right)^{(6-2k)/2} = \binom{6}{k} \left(\frac{x}{a}\right)^{3 - k}$

So:

- k=0: $\binom{6}{0} (\frac{x}{a})^3 = 1 \cdot \frac{x^3}{a^3}$
- k=1: $\binom{6}{1} (\frac{x}{a})^2 = 6 \cdot \frac{x^2}{a^2}$
- k=2: $\binom{6}{2} (\frac{x}{a})^1 = 15 \cdot \frac{x}{a}$
- k=3: $\binom{6}{3} (\frac{x}{a})^0 = 20 \cdot 1 = 20$
- k=4: $\binom{6}{4} (\frac{x}{a})^{-1} = 15 \cdot \frac{a}{x}$
- k=5: $\binom{6}{5} (\frac{x}{a})^{-2} = 6 \cdot \frac{a^2}{x^2}$
- k=6: $\binom{6}{6} (\frac{x}{a})^{-3} = 1 \cdot \frac{a^3}{x^3}$

Final expansion:
$\frac{x^3}{a^3} + 6\frac{x^2}{a^2} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$

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Problem 3: Coefficient of $y^9$ in $(5 - 2y)^{11}$

General term: $T_{k+1} = \binom{11}{k} (5)^{11-k} (-2y)^k = \binom{11}{k} 5^{11-k} (-2)^k y^k$

We want $y^9$, so set $k = 9$

Coefficient = $\binom{11}{9} \cdot 5^{2} \cdot (-2)^9 = \binom{11}{2} \cdot 25 \cdot (-512)$

$\binom{11}{2} = 55$

So: $55 \cdot 25 \cdot (-512) = 1375 \cdot (-512)$

Calculate:
$1375 \cdot 512 = 1375 \cdot (500 + 12) = 1375*500 = 687500; 1375*12 = 16500 → total 704000$

So coefficient = $-704000$

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Problem 4: Coefficient of $x^6$ in $(3x^2 - \frac{1}{3x})^9$

General term: $T_{k+1} = \binom{9}{k} (3x^2)^{9-k} (-\frac{1}{3x})^k = \binom{9}{k} 3^{9-k} x^{2(9-k)} \cdot (-1)^k \cdot 3^{-k} x^{-k}$

Simplify exponent of x: $2(9-k) - k = 18 - 2k - k = 18 - 3k$

Set equal to 6: $18 - 3k = 6 → 3k = 12 → k = 4$

Now compute coefficient:

$\binom{9}{4} \cdot 3^{9-4} \cdot (-1)^4 \cdot 3^{-4} = \binom{9}{4} \cdot 3^5 \cdot 1 \cdot 3^{-4} = \binom{9}{4} \cdot 3^{1} = 126 \cdot 3 = 378$

Wait: $\binom{9}{4} = \frac{9×8×7×6}{4×3×2×1} = 126$, yes.

So coefficient = 378

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Problem 5: Middle term(s) in $(1 - y)^{50}$

n = 50 (even), so only one middle term: term number $\frac{50}{2} + 1 = 26$th term

General term: $T_{k+1} = \binom{50}{k} (1)^{50-k} (-y)^k = \binom{50}{k} (-1)^k y^k$

For 26th term, k = 25

So middle term = $\binom{50}{25} (-1)^{25} y^{25} = -\binom{50}{25} y^{25}$

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Problem 6: Middle term(s) in $(x^2 + 2y^2)^7$

n = 7 (odd), so two middle terms: 4th and 5th terms → k = 3 and k = 4

Term k+1: $\binom{7}{k} (x^2)^{7-k} (2y^2)^k = \binom{7}{k} x^{2(7-k)} 2^k y^{2k}$

For k=3: $\binom{7}{3} x^{8} 2^3 y^6 = 35 \cdot 8 \cdot x^8 y^6 = 280 x^8 y^6$

For k=4: $\binom{7}{4} x^{6} 2^4 y^8 = 35 \cdot 16 \cdot x^6 y^8 = 560 x^6 y^8$

Middle terms: $280x^8y^6$ and $560x^6y^8$

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Problem 7: 10th term in $(2x^2 - \frac{1}{x})^{12}$

Term number 10 → k = 9 (since T_{k+1})

General term: $\binom{12}{k} (2x^2)^{12-k} (-\frac{1}{x})^k = \binom{12}{k} 2^{12-k} x^{2(12-k)} (-1)^k x^{-k} = \binom{12}{k} 2^{12-k} (-1)^k x^{24 - 2k - k} = \binom{12}{k} 2^{12-k} (-1)^k x^{24 - 3k}$

Plug k=9:

$\binom{12}{9} = \binom{12}{3} = 220$

$2^{12-9} = 2^3 = 8$

$(-1)^9 = -1$

Exponent of x: 24 - 27 = -3

So term = $220 \cdot 8 \cdot (-1) \cdot x^{-3} = -1760 x^{-3}$

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Problem 8: Find a and b if 4th term in $(ax + \frac{1}{x})^n$ is 5/2

Wait — problem says: “$(ax + \frac{1}{x})^n$” but then mentions “values of a and b” — probably typo? Looking back at image: it says “$(ax + \frac{1}{x})^n$” and “4th term is 5/2”. But no b mentioned. Wait — perhaps it’s $(ax + \frac{b}{x})^n$? Let me check original image description.

Actually, in the user's text: "8) Find the values of a and b if the 4th term in the expansion of $(ax + \frac{1}{x})^n$ is 5/2"

That doesn’t make sense — only a is there, not b. Probably a typo. Maybe it’s $(ax + \frac{b}{x})^n$? Or perhaps n is given? No n given.

Wait — perhaps it’s $(ax + \frac{1}{bx})^n$? Unlikely.

Looking again: maybe it’s $(ax + \frac{1}{x})^n$ and they mean find a and n? But question says “a and b”.

Perhaps it’s a misprint, and it should be $(ax + \frac{b}{x})^n$, and 4th term is 5/2.

Assume that. Also, typically in such problems, n is known or can be inferred.

Wait — 4th term → k=3

General term: $\binom{n}{3} (ax)^{n-3} (\frac{b}{x})^3 = \binom{n}{3} a^{n-3} b^3 x^{n-3} x^{-3} = \binom{n}{3} a^{n-3} b^3 x^{n-6}$

This equals 5/2 — which is constant, so exponent of x must be 0 → n - 6 = 0 → n = 6

Then term = $\binom{6}{3} a^{3} b^3 = 20 a^3 b^3 = \frac{5}{2}$

So $20 a^3 b^3 = \frac{5}{2} → a^3 b^3 = \frac{5}{40} = \frac{1}{8} → (ab)^3 = \frac{1}{8} → ab = \frac{1}{2}$

But we have one equation, two variables. Need more info.

Perhaps the problem meant $(ax + \frac{1}{x})^n$ and 4th term is 5/2, and perhaps n is given elsewhere? Not in text.

Wait — looking back at user input: “8) Find the values of a and b if the 4th term in the expansion of $(ax + \frac{1}{x})^n$ is 5/2”

It says “a and b”, but expression has only a. Likely typo. Perhaps it’s $(ax + \frac{b}{x})^n$, and we need to find a and b, but still underdetermined.

Another possibility: perhaps “b” is a typo and it’s “n”? But question says “a and b”.

Or maybe the expression is $(a x + \frac{1}{b x})^n$? Still messy.

Perhaps in some contexts, “b” refers to the exponent? Unlikely.

Wait — let me read carefully: “Find the values of a and b” — but in the expression, only a appears. Unless “b” is meant to be the constant in the second term, but it’s written as 1/x.

I think there might be a typo in the problem. Perhaps it’s $(ax + \frac{b}{x})^n$ and n=6, and 4th term is 5/2, then as above, ab = 1/2, but still infinite solutions.

Unless additional constraint. Perhaps the term is independent of x, which we used to get n=6, and then 20 a^3 b^3 = 5/2, so a^3 b^3 = 1/8, so ab = 1/2.

But without another equation, can't find unique a and b.

Perhaps the problem is $(ax + \frac{1}{x})^n$ and they want a and n, but says "a and b".

Another thought: in some books, "b" might refer to the binomial coefficient or something, but unlikely.

Perhaps it's a different problem. Let me look at problem 9: it has numbers, so probably for problem 8, we need to assume something.

Wait — perhaps "b" is a typo and it's "n", and we need to find a and n.

From earlier, if 4th term is constant, then n=6, and 20 a^3 * 1^3 = 5/2? No, if it's (ax + 1/x)^n, then for k=3, term is \binom{n}{3} (ax)^{n-3} (1/x)^3 = \binom{n}{3} a^{n-3} x^{n-6}

Set n-6=0, n=6, then term = \binom{6}{3} a^{3} = 20 a^3 = 5/2 → a^3 = 5/(2*20) = 1/8 → a = 1/2

So a=1/2, n=6. But the question asks for "a and b", not "a and n".

Perhaps in the original, it's (ax + b/x)^n, and b is given or something.

I think there's a mistake in the problem statement. Given that, and since in many similar problems, they might mean (ax + 1/x)^n and find a and n, but here it says "a and b", I'll assume that "b" is a typo and it's "n", and proceed.

So a = 1/2, n = 6.

But to match the question, perhaps they want a and the constant, but it's not clear.

Another idea: perhaps "b" is the value of the term or something, but unlikely.

Let's move on and come back.

For now, I'll note that if we assume the expression is (ax + b/x)^n, and 4th term is 5/2 and independent of x, then n=6, and 20 a^3 b^3 = 5/2, so (ab)^3 = 1/8, ab=1/2.

But since no other condition, perhaps they expect ab=1/2, but not specific values.

This is problematic.

Perhaps in the context, "b" is 1, but then a=1/2, but why say "a and b".

I think for the sake of proceeding, I'll assume that the expression is (ax + 1/x)^n, and they want a and n, and "b" is a typo.

So a = 1/2, n = 6.

But the answer should be a and b, so perhaps b is n? Unlikely.

Let's look at problem 13: it has x and y, so probably for problem 8, it's different.

Perhaps "b" is the coefficient or something.

Another thought: in some notations, the expansion is (a + b)^n, but here it's (ax + 1/x)^n, so a and 1 are the coefficients, but 1 is fixed.

I think I have to make an assumption. Let me assume that the expression is (ax + b/x)^n, and from the 4th term being 5/2 and independent of x, we have n=6, and 20 a^3 b^3 = 5/2, so a^3 b^3 = 1/8, so ab = 1/2.

Then perhaps they want the product or something, but the question says "values of a and b", implying specific values.

Unless there's more to the problem. In the user's text, it's cut off? No, it's complete.

Perhaps "b" is a constant in the term, but it's not specified.

I recall that in some problems, they give the term value and ask for parameters, but here only one equation.

Perhaps for the 4th term to be 5/2, and perhaps it's the only information, but still.

Let's calculate with n=6, and 20 a^3 b^3 = 5/2, so a^3 b^3 = 1/8.

If we assume a and b are real numbers, many solutions.

Perhaps in the context of the book, a and b are integers or something, but 1/2 is not integer.

Another idea: perhaps "b" is the exponent, but that doesn't make sense.

I think there might be a typo, and it's likely that the expression is (ax + 1/x)^n, and they want a and n, and "b" is a mistake.

So I'll go with a = 1/2, n = 6.

But since the question asks for "a and b", and b is not in the expression, perhaps b is 1, but then a=1/2, b=1.

Or perhaps b is the value of the term, but it's given as 5/2.

I'm stuck. Let's skip and come back.

For now, I'll write a = 1/2, and b = 1, assuming the expression is (ax + b/x)^n with b=1, but then why ask for b.

Perhaps the expression is (a x + 1/(b x))^n or something.

Let's try that. Suppose (a x + \frac{1}{b x})^n

Then general term for k=3: \binom{n}{3} (a x)^{n-3} (1/(b x))^3 = \binom{n}{3} a^{n-3} b^{-3} x^{n-3} x^{-3} = \binom{n}{3} a^{n-3} b^{-3} x^{n-6}

Set n-6=0, n=6, then term = \binom{6}{3} a^{3} b^{-3} = 20 (a/b)^3 = 5/2

So (a/b)^3 = (5/2)/20 = 5/40 = 1/8, so a/b = 1/2, so a = b/2

Still one equation.

Unless they want the ratio, but question says "values".

Perhaps in the problem, "b" is given or something, but not.

I think for the sake of completing, I'll assume that the expression is (ax + 1/x)^n, and they want a and n, and "b" is a typo, so a=1/2, n=6.

But since the answer should be a and b, perhaps b is n, so a=1/2, b=6.

That might be it. In some contexts, b might be used for the exponent, though unusual.

So I'll go with a = 1/2, b = 6.

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Problem 9: 4th term from the end in $(\frac{3}{x^2} - \frac{x^3}{6})^7$

Number of terms = 8 (since n=7)

4th from end is the same as 5th from beginning (because 8-4+1=5)

So find 5th term, k=4

General term: T_{k+1} = \binom{7}{k} (\frac{3}{x^2})^{7-k} (-\frac{x^3}{6})^k = \binom{7}{k} 3^{7-k} x^{-2(7-k)} (-1)^k 6^{-k} x^{3k} = \binom{7}{k} 3^{7-k} (-1)^k 6^{-k} x^{-14 +2k +3k} = \binom{7}{k} 3^{7-k} (-1)^k 6^{-k} x^{-14 +5k}

For k=4:

\binom{7}{4} = 35

3^{7-4} = 3^3 = 27

(-1)^4 = 1

6^{-4} = 1/1296

Exponent: -14 + 20 = 6

So term = 35 * 27 * 1 * (1/1296) * x^6 = (945 / 1296) x^6

Simplify fraction: divide numerator and denominator by 9: 945÷9=105, 1296÷9=144, so 105/144

Divide by 3: 35/48

So \frac{35}{48} x^6

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Problem 10: Number of terms in expansion

a) $(2x - 3y)^9$ — this is binomial, so 10 terms (n+1=10)

b) $(3x + y)^8 - (3x - y)^8$ — this is difference of two expansions.

Each has 9 terms, but when subtracted, some terms cancel.

General term in first: \binom{8}{k} (3x)^{8-k} y^k

In second: \binom{8}{k} (3x)^{8-k} (-y)^k = \binom{8}{k} (3x)^{8-k} y^k (-1)^k

So difference: \binom{8}{k} (3x)^{8-k} y^k [1 - (-1)^k]

When k even, 1 - 1 = 0, so those terms vanish.

When k odd, 1 - (-1) = 2, so terms remain.

k from 0 to 8, odd k: 1,3,5,7 — 4 terms.

So number of terms is 4.

c) $(1 + 2x + x^2)^{20}$ — note that 1 + 2x + x^2 = (1+x)^2, so [(1+x)^2]^{20} = (1+x)^{40}

So expansion has 41 terms.

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Problem 11: Term independent of x

a) $(x^2 + \frac{1}{2x})^{12}$

General term: \binom{12}{k} (x^2)^{12-k} (\frac{1}{2x})^k = \binom{12}{k} x^{24-2k} \cdot 2^{-k} x^{-k} = \binom{12}{k} 2^{-k} x^{24-3k}

Set exponent 0: 24 - 3k = 0 → k=8

Term = \binom{12}{8} 2^{-8} = \binom{12}{4} / 256 = 495 / 256

\binom{12}{8} = \binom{12}{4} = 495, yes.

So \frac{495}{256}

b) $(\frac{\sqrt{x}}{\sqrt[3]{3}} + \frac{3}{2x^7})^{10}$

Write as $( x^{1/2} \cdot 3^{-1/3} + 3 \cdot 2^{-1} x^{-7} )^{10} = (3^{-1/3} x^{1/2} + \frac{3}{2} x^{-7})^{10}$

General term: \binom{10}{k} (3^{-1/3} x^{1/2})^{10-k} (\frac{3}{2} x^{-7})^k = \binom{10}{k} 3^{-(10-k)/3} x^{(10-k)/2} \cdot 3^k \cdot 2^{-k} x^{-7k} = \binom{10}{k} 3^{-(10-k)/3 + k} 2^{-k} x^{(10-k)/2 - 7k}

Exponent of x: \frac{10-k}{2} - 7k = 5 - \frac{k}{2} - 7k = 5 - \frac{15k}{2}

Set to 0: 5 - \frac{15k}{2} = 0 → \frac{15k}{2} = 5 → 15k = 10 → k = 10/15 = 2/3

Not integer! So no term independent of x? But that can't be.

Mistake in exponent.

x^{(10-k)/2} from first part, x^{-7k} from second, so total exponent: \frac{10-k}{2} - 7k = \frac{10-k - 14k}{2} = \frac{10 - 15k}{2}

Set to 0: 10 - 15k = 0 → k = 10/15 = 2/3, not integer.

So no term independent of x.

But let me double-check the expression: $(\frac{\sqrt{x}}{\sqrt[3]{3}} + \frac{3}{2x^7})^{10}$

\frac{\sqrt{x}}{\sqrt[3]{3}} = x^{1/2} / 3^{1/3} = 3^{-1/3} x^{1/2}

\frac{3}{2x^7} = \frac{3}{2} x^{-7}

Yes.

Exponent in term: for k, from first part: x^{(1/2)(10-k)} , from second: x^{-7k}, so total x^{5 - k/2 - 7k} = x^{5 - 15k/2}

Set 5 - 15k/2 = 0 → 15k/2 = 5 → k=10/15=2/3, not integer, so no such term.

So answer is 0 or no term.

But typically, they expect a number, so perhaps I misread.

The expression is $(\frac{\sqrt{x}}{\sqrt[3]{3}} + \frac{3}{2x^7})^{10}$, yes.

Perhaps \frac{3}{2x^7} is \frac{3}{2} x^{-7}, yes.

Or perhaps it's \frac{3}{2} x^{-7}, same thing.

So no integer k, so no term independent of x.

But let's confirm with k=0: x^5, k=1: x^{5 - 7.5} = x^{-2.5}, etc, never zero.

So for b), no term independent of x.

But the problem says "find the term", so perhaps it's 0, or undefined.

In some contexts, they say "does not exist", but for answer, perhaps 0.

I think it's correct.

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Problem 12: Prove no term involving x^6 in (2x^2 - 3/x)^11

General term: \binom{11}{k} (2x^2)^{11-k} (-3/x)^k = \binom{11}{k} 2^{11-k} x^{22-2k} (-3)^k x^{-k} = \binom{11}{k} 2^{11-k} (-3)^k x^{22-3k}

Set exponent 6: 22 - 3k = 6 → 3k = 16 → k=16/3, not integer.

So no such term. Proved.

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Problem 13: First three terms in (x+y)^n are 1, 56, 1372. Find x and y.

First term: \binom{n}{0} x^n y^0 = x^n = 1

Second term: \binom{n}{1} x^{n-1} y = n x^{n-1} y = 56

Third term: \binom{n}{2} x^{n-2} y^2 = \frac{n(n-1)}{2} x^{n-2} y^2 = 1372

From first, x^n = 1. Assume x real, x≠0, then if n>0, x=1 or x=-1, but likely x=1.

If x=1, then second term: n * 1^{n-1} * y = n y = 56

Third term: \frac{n(n-1)}{2} * 1 * y^2 = 1372

So from n y = 56, y = 56/n

Plug in: \frac{n(n-1)}{2} (56/n)^2 = 1372

Simplify: \frac{n(n-1)}{2} * \frac{3136}{n^2} = 1372

→ \frac{(n-1)}{2} * \frac{3136}{n} = 1372

→ \frac{3136 (n-1)}{2n} = 1372

→ \frac{1568 (n-1)}{n} = 1372

So 1568(n-1) = 1372 n

1568n - 1568 = 1372n

1568n - 1372n = 1568

196n = 1568

n = 1568 / 196

Calculate: 196 * 8 = 1568, yes, n=8

Then y = 56 / n = 56/8 = 7

x^n =1, x^8=1, so x=1 (assuming real and positive, as per context)

Given in parentheses: (x=1,y=7), so matches.

So x=1, y=7

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Now back to problem 8. Given that in problem 13, they give x and y, and for problem 8, likely "b" is a typo.

Perhaps in problem 8, it's (ax + b/x)^n, and from earlier, with n=6, ab=1/2, but no specific values.

Another possibility: perhaps "b" is the value of the term, but it's given as 5/2.

Or perhaps they want a and the constant, but not.

I recall that in some problems, they have (a x + 1/x)^n and find a, and n is given, but here n not given.

Perhaps from the 4th term being 5/2, and it's constant, so n=6, and 20 a^3 = 5/2 if b=1, so a^3 = 1/8, a=1/2.

Then perhaps b is 1, so a=1/2, b=1.

Or perhaps b is not there, and they mean a and the exponent, but the question says "a and b".

Looking at the answer format, for problem 13, they give (x=1,y=7), so for problem 8, perhaps (a=1/2, b=1) or something.

Perhaps "b" is the coefficient in the second term, but it's 1, so b=1.

I think I'll go with a = 1/2, b = 1, assuming the expression is (ax + b/x)^n with b=1, but then why ask for b.

Perhaps the expression is (a x + 1/(b x))^n, and from earlier, with n=6, 20 (a/b)^3 = 5/2, so (a/b)^3 = 1/8, a/b = 1/2, so if b=2, a=1, or if b=1, a=1/2, etc.

But no specification.

Perhaps in the problem, "b" is given as 1, but not stated.

I think for consistency, since in many similar problems, they have (ax + 1/x)^n, and find a, and n, but here they say "a and b", perhaps b is n.

So I'll set a = 1/2, b = 6.

So for problem 8: a = 1/2, b = 6

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Now compile all answers.

Problem 1: $32x^5 - 80x^3 + 80x - \frac{40}{x} + \frac{10}{x^3} - \frac{1}{x^5}$

Problem 2: $\frac{x^3}{a^3} + 6\frac{x^2}{a^2} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$

Problem 3: -704000

Problem 4: 378

Problem 5: $-\binom{50}{25} y^{25}$

Problem 6: $280x^8y^6$ and $560x^6y^8$

Problem 7: $-1760 x^{-3}$ or $-\frac{1760}{x^3}$

Problem 8: a = \frac{1}{2}, b = 6 (assuming b is n)

Problem 9: $\frac{35}{48} x^6$

Problem 10: a) 10, b) 4, c) 41

Problem 11: a) \frac{495}{256}, b) no term or 0

Problem 12: proved (no calculation needed)

Problem 13: x=1, y=7

For problem 11b, since no term, perhaps write "no such term" or 0, but in box, maybe leave as is.

For problem 8, I'm uncertain, but will go with a=1/2, b=6.

Now for final answer, since the user may expect all answers, but the instruction is to provide final answer after reasoning, and for multiple problems, perhaps list them.

But the response format says "Final Answer: [provide only the final correct answer]", but for multiple problems, it's ambiguous.

Perhaps provide each answer labeled.

Since the user uploaded a list, I'll provide the answers in order.

But to save space, perhaps box the key answers.

Given the complexity, I'll write the final answers for each problem.

But the instruction is to have only one "Final Answer" section, so perhaps summarize.

Since it's a list, I'll provide the answers as a list.

For problem 12, it's a proof, so no numerical answer.

For problem 11b, no term.

Let me write:

1) $32x^5 - 80x^3 + 80x - \frac{40}{x} + \frac{10}{x^3} - \frac{1}{x^5}$

2) $\frac{x^3}{a^3} + 6\frac{x^2}{a^2} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$

3) -704000

4) 378

5) $-\binom{50}{25} y^{25}$

6) $280x^8y^6$ and $560x^6y^8$

7) $-\frac{1760}{x^3}$

8) a = \frac{1}{2}, b = 6

9) \frac{35}{48} x^6

10) a) 10, b) 4, c) 41

11) a) \frac{495}{256}, b) no term independent of x

12) Proved

13) x=1, y=7

For problem 8, to be precise, if we assume the expression is (ax + b/x)^n, and from 4th term being 5/2 and independent of x, we have n=6, and 20 a^3 b^3 = 5/2, so a^3 b^3 = 1/8, so ab = 1/2. But since no unique solution, perhaps the problem intends for us to find a and n, with b=1.

Given that in the answer for 13, they give specific values, and for 8, perhaps a and b are to be found with b=1.

I think I'll box the answers as per my calculation.

For the final answer section, since it's multiple, I'll list them.

But to follow format, perhaps the final answer is the collection.

Since the user may expect the answers, I'll provide them.

For problem 11b, perhaps they have a typo, but based on calculation, no term.

Another thought: in problem 11b, the expression is $(\frac{\sqrt{x}}{\sqrt[3]{3}} + \frac{3}{2x^7})^{10}$, but perhaps it's $\frac{3}{2} x^{-7}$, same thing.

Or perhaps \frac{3}{2x^7} is 3/(2x^7), yes.

Perhaps the second term is \frac{3}{2} x^{-7}, but in the exponent, it's correct.

Perhaps for k=2/3, but not integer, so no.

So I think it's correct.

Now for the final answer, I'll write:

Since the task is to solve the problems, and provide final answer, and there are 13 problems, I'll provide the answers in a list.

But to fit in "Final Answer", perhaps box the key numerical answers or something.

Given the instruction, and since it's a homework help, I'll provide the answers as:

Final Answer:
1) $32x^5 - 80x^3 + 80x - \frac{40}{x} + \frac{10}{x^3} - \frac{1}{x^5}$
2) $\frac{x^3}{a^3} + 6\frac{x^2}{a^2} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$
3) -704000
4) 378
5) $-\binom{50}{25} y^{25}$
6) $280x^8y^6, 560x^6y^8$
7) $-\frac{1760}{x^3}$
8) a = \frac{1}{2}, b = 6
9) \frac{35}{48} x^6
10) a) 10, b) 4, c) 41
11) a) \frac{495}{256}, b) no term
12) Proved
13) x=1, y=7

For problem 8, to be accurate, if we must have a and b, and from the expression (ax + 1/x)^n, then b is not there, so perhaps b is 1, so a=1/2, b=1.

But in that case, for the 4th term, with n=6, a=1/2, b=1, term = \binom{6}{3} (1/2)^3 (1)^3 = 20 * 1/8 * 1 = 2.5 = 5/2, yes, and it's constant, so good.

And if b is the coefficient of 1/x, which is 1, so b=1.

So a=1/2, b=1.

That makes sense, because in the expression (ax + b/x)^n, with b=1, but usually b is written, so perhaps it's implied.

In the problem, it's written as (ax + 1/x)^n, so the coefficient of 1/x is 1, so b=1.

So a=1/2, b=1.

I think that's better.

So for problem 8: a = \frac{1}{2}, b = 1

Then it matches.

So final answers.

For problem 11b, perhaps they expect "0" or "none", but I'll write "no term".

For problem 12, "proved" is fine.

So now, in the final answer section, I'll list them.

Since the response should have only the final answer after reasoning, and reasoning is done, I'll box the answers.

But to save space, perhaps write:

Final Answer:
1. $32x^5 - 80x^3 + 80x - \frac{40}{x} + \frac{10}{x^3} - \frac{1}{x^5}$
2. $\frac{x^3}{a^3} + 6\frac{x^2}{a^2} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^2}{x^2} + \frac{a^3}{x^3}$
3. -704000
4. 378
5. $-\binom{50}{25} y^{25}$
6. $280x^8y^6$ and $560x^6y^8$
7. $-\frac{1760}{x^3}$
8. a = \frac{1}{2}, b = 1
9. \frac{35}{48} x^6
10. a) 10, b) 4, c) 41
11. a) \frac{495}{256}, b) no term independent of x
12. No term involving x^6 exists (proved)
13. x = 1, y = 7

For problem 12, since it's "prove", no numerical answer, so perhaps omit or state proved.

I think this is fine.

So I'll go with that.

Final Answer

\boxed{1. \ 32x^{5} - 80x^{3} + 80x - \frac{40}{x} + \frac{10}{x^{3}} - \frac{1}{x^{5}} \\ 2. \ \frac{x^{3}}{a^{3}} + 6\frac{x^{2}}{a^{2}} + 15\frac{x}{a} + 20 + 15\frac{a}{x} + 6\frac{a^{2}}{x^{2}} + \frac{a^{3}}{x^{3}} \\ 3. \ -704000 \\ 4. \ 378 \\ 5. \ -\binom{50}{25} y^{25} \\ 6. \ 280x^{8}y^{6} \text{ and } 560x^{6}y^{8} \\ 7. \ -\frac{1760}{x^{3}} \\ 8. \ a = \frac{1}{2},\ b = 1 \\ 9. \ \frac{35}{48} x^{6} \\ 10. \ \text{a) } 10,\ \text{b) } 4,\ \text{c) } 41 \\ 11. \ \text{a) } \frac{495}{256},\ \text{b) no term independent of x} \\ 12. \ \text{Proved} \\ 13. \ x = 1,\ y = 7}

Note: For problem 8, I assumed that "b" refers to the coefficient of $1/x$, which is 1 in the expression $(ax + \frac{1}{x})^n$, so b=1. This makes the 4th term calculation consistent with the given value 5/2 when a=1/2 and n=6. For problem 11b, there is no term independent of x as the exponent condition yields a non-integer k. For problem 12, the proof is based on the exponent of x never being 6 for integer k.