CBSE Class 11 Principle of Mathematical Induction Worksheet A - Free Printable

Let's solve each of these problems one by one, using mathematical reasoning and proofs (mostly mathematical induction where appropriate). We'll go step-by-step.

---

1. For every integer $ n $, prove that $ 7^n - 3^n $ is divisible by 4.

We will use mathematical induction.

#### Base case: $ n = 0 $
$$
7^0 - 3^0 = 1 - 1 = 0 \quad \text{and } 0 \div 4 = 0 \Rightarrow \text{Divisible by 4.}
$$

#### Inductive step:
Assume for some $ k \geq 0 $, $ 7^k - 3^k $ is divisible by 4, i.e.,
$$
7^k - 3^k \equiv 0 \pmod{4} \Rightarrow 7^k \equiv 3^k \pmod{4}
$$

Now consider $ n = k+1 $:
$$
7^{k+1} - 3^{k+1} = 7 \cdot 7^k - 3 \cdot 3^k
= 7 \cdot 7^k - 7 \cdot 3^k + 7 \cdot 3^k - 3 \cdot 3^k
= 7(7^k - 3^k) + 3^k(7 - 3)
= 7(7^k - 3^k) + 4 \cdot 3^k
$$

- By inductive hypothesis: $ 7^k - 3^k \equiv 0 \pmod{4} $ → so $ 7(7^k - 3^k) \equiv 0 \pmod{4} $
- $ 4 \cdot 3^k \equiv 0 \pmod{4} $

So the whole expression is divisible by 4.

✔ Hence, $ 7^n - 3^n $ is divisible by 4 for all integers $ n \geq 0 $.

> Note: For negative $ n $, this may not hold since $ 7^{-n}, 3^{-n} $ are fractions. But usually such problems assume $ n \in \mathbb{N}_0 $. So we assume $ n \geq 0 $.

---

2. Prove that $ n(n+1)(n+5) $ is a multiple of 3.

We analyze modulo 3.

Any integer $ n $ modulo 3 is either $ 0, 1, $ or $ 2 $.

Check all cases:

- If $ n \equiv 0 \pmod{3} $ → $ n $ divisible by 3 → product divisible by 3.
- If $ n \equiv 1 \pmod{3} $ → $ n+2 \equiv 3 \equiv 0 $? Wait: $ n = 1 $ → $ n+5 = 6 \equiv 0 \pmod{3} $
- So $ n+5 \equiv 0 \pmod{3} $
- If $ n \equiv 2 \pmod{3} $ → $ n+1 = 3 \equiv 0 \pmod{3} $

Thus, in all cases, one of $ n, n+1, n+5 $ is divisible by 3.

✔ Therefore, $ n(n+1)(n+5) $ is divisible by 3 for all integers $ n $.

---

3. Prove that $ 10^{2n+1} + 1 $ is divisible by 11.

Use induction or modular arithmetic.

Let’s try modulo 11.

Note: $ 10 \equiv -1 \pmod{11} $

Then:
$$
10^{2n+1} \equiv (-1)^{2n+1} \pmod{11}
\Rightarrow (-1)^{\text{odd}} = -1
\Rightarrow 10^{2n+1} \equiv -1 \pmod{11}
\Rightarrow 10^{2n+1} + 1 \equiv -1 + 1 = 0 \pmod{11}
$$

✔ So $ 10^{2n+1} + 1 $ is divisible by 11 for all $ n \geq 0 $.

---

4. Prove $ \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{n}\right) = n+1 $

Left-hand side:
$$
\prod_{k=1}^n \left(1 + \frac{1}{k}\right) = \prod_{k=1}^n \frac{k+1}{k}
$$

This is a telescoping product:
$$
\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{n+1}{n}
$$

All intermediate terms cancel:
- Numerator: $ 2 \cdot 3 \cdot 4 \cdots (n+1) $
- Denominator: $ 1 \cdot 2 \cdot 3 \cdots n $

So result = $ \frac{n+1}{1} = n+1 $

✔ Proved.

---

5. Prove $ 1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3} $

Let $ S_n = \sum_{k=1}^n k(k+1) $

We know:
$$
k(k+1) = k^2 + k
\Rightarrow S_n = \sum_{k=1}^n (k^2 + k) = \sum k^2 + \sum k
= \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}
$$

Factor $ \frac{n(n+1)}{2} $:
$$
= \frac{n(n+1)}{2} \left( \frac{2n+1}{3} + 1 \right)
= \frac{n(n+1)}{2} \left( \frac{2n+1 + 3}{3} \right)
= \frac{n(n+1)}{2} \cdot \frac{2n+4}{3}
= \frac{n(n+1)}{2} \cdot \frac{2(n+2)}{3}
= \frac{n(n+1)(n+2)}{3}
$$

✔ Proved.

---

6. Prove $ (2n+7) < (n+3)^2 $ for all $ n \geq 0 $?

Wait — the inequality is likely meant to be for all sufficiently large $ n $, or perhaps it's supposed to be true for all $ n \geq 0 $?

Let’s test small values:

- $ n = 0 $: $ 2(0)+7 = 7 $, $ (0+3)^2 = 9 $ → $ 7 < 9 $ ✔
- $ n = 1 $: $ 9 < 16 $ ✔
- $ n = 2 $: $ 11 < 25 $ ✔
- $ n = 3 $: $ 13 < 36 $ ✔

But let’s prove it algebraically.

We want:
$$
2n + 7 < (n+3)^2 = n^2 + 6n + 9
\Rightarrow 0 < n^2 + 6n + 9 - 2n - 7 = n^2 + 4n + 2
$$

Now $ n^2 + 4n + 2 > 0 $ for all $ n \geq 0 $, since minimum at $ n = -2 $, and increasing.

For $ n \geq 0 $, clearly positive.

✔ So $ (2n+7) < (n+3)^2 $ for all $ n \geq 0 $.

---

7. Prove $ \frac{1}{1\cdot4} + \frac{1}{4\cdot7} + \frac{1}{7\cdot10} + \cdots + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1} $

This is a telescoping series.

General term: $ \frac{1}{(3k-2)(3k+1)} $

Use partial fractions:
Let $ \frac{1}{(3k-2)(3k+1)} = \frac{A}{3k-2} + \frac{B}{3k+1} $

Multiply both sides:
$$
1 = A(3k+1) + B(3k-2)
\Rightarrow 1 = (3A + 3B)k + (A - 2B)
$$

Match coefficients:
- $ 3A + 3B = 0 \Rightarrow A + B = 0 $
- $ A - 2B = 1 $

From $ A = -B $, plug in:
$ -B - 2B = 1 \Rightarrow -3B = 1 \Rightarrow B = -\frac{1}{3}, A = \frac{1}{3} $

So:
$$
\frac{1}{(3k-2)(3k+1)} = \frac{1}{3} \left( \frac{1}{3k-2} - \frac{1}{3k+1} \right)
$$

Sum from $ k=1 $ to $ n $:
$$
\sum_{k=1}^n \frac{1}{(3k-2)(3k+1)} = \frac{1}{3} \sum_{k=1}^n \left( \frac{1}{3k-2} - \frac{1}{3k+1} \right)
$$

Write out terms:
$$
\frac{1}{3} \left[ \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \cdots + \left(\frac{1}{3n-2} - \frac{1}{3n+1}\right) \right]
$$

Telescopes: all intermediate terms cancel.

Result:
$$
\frac{1}{3} \left(1 - \frac{1}{3n+1}\right) = \frac{1}{3} \cdot \frac{3n+1 - 1}{3n+1} = \frac{1}{3} \cdot \frac{3n}{3n+1} = \frac{n}{3n+1}
$$

✔ Proved.

---

8. Prove $ 1\cdot2 + 2\cdot2^2 + 3\cdot2^3 + \cdots + n\cdot2^n = (n-1)2^{n+1} + 2 $

Let $ S_n = \sum_{k=1}^n k \cdot 2^k $

We use induction.

#### Base case: $ n = 1 $
LHS: $ 1 \cdot 2^1 = 2 $

RHS: $ (1-1)2^{2} + 2 = 0 + 2 = 2 $ ✔

#### Inductive step:
Assume true for $ n = k $:
$$
\sum_{i=1}^k i \cdot 2^i = (k-1)2^{k+1} + 2
$$

Now for $ n = k+1 $:
$$
S_{k+1} = S_k + (k+1) \cdot 2^{k+1}
= (k-1)2^{k+1} + 2 + (k+1) \cdot 2^{k+1}
= \left[(k-1) + (k+1)\right] 2^{k+1} + 2
= (2k) \cdot 2^{k+1} + 2 = k \cdot 2^{k+2} + 2
$$

Now check RHS for $ n = k+1 $:
$$
((k+1)-1)2^{(k+1)+1} + 2 = k \cdot 2^{k+2} + 2
$$

Matches!

✔ So by induction, formula holds for all $ n \geq 1 $.

---

## ✔ Summary of All Proofs:

| Problem | Method | Result |
|--------|--------|--------|
| 1 | Induction / Modulo | $ 7^n - 3^n \equiv 0 \pmod{4} $ |
| 2 | Case analysis mod 3 | $ n(n+1)(n+5) \equiv 0 \pmod{3} $ |
| 3 | Modulo 11 | $ 10^{2n+1} + 1 \equiv 0 \pmod{11} $ |
| 4 | Telescoping product | Product = $ n+1 $ |
| 5 | Summation formulas | $ \sum k(k+1) = \frac{n(n+1)(n+2)}{3} $ |
| 6 | Algebraic inequality | $ 2n+7 < (n+3)^2 $ for $ n \geq 0 $ |
| 7 | Partial fractions & telescoping | Sum = $ \frac{n}{3n+1} $ |
| 8 | Mathematical induction | $ \sum k \cdot 2^k = (n-1)2^{n+1} + 2 $ |

Let me know if you'd like any of these explained further!