Worksheet on recursive sequences with problems and a scoring table.

It looks like you've uploaded a worksheet with several problems related to sequences and series, specifically asking for recursive formulas, explicit formulas, and finding terms of sequences.

Since I can't view images directly, I'll assume the content is as described in your text — a math worksheet with 10 problems involving sequences. Below, I’ll solve each problem step by step based on typical sequence patterns and standard notation.

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Problem 1:

Write a recursive formula for the sequence:
4, 12, 30, 60, ...

Let’s examine the pattern:

- $ a_1 = 4 $
- $ a_2 = 12 $
- $ a_3 = 30 $
- $ a_4 = 60 $

Look at the ratios or differences:

- $ 12 / 4 = 3 $
- $ 30 / 12 = 2.5 $
- $ 60 / 30 = 2 $ → not constant ratio

Now check differences:
- $ 12 - 4 = 8 $
- $ 30 - 12 = 18 $
- $ 60 - 30 = 30 $

Second differences:
- $ 18 - 8 = 10 $
- $ 30 - 18 = 12 $ → not constant

Try factoring:
- $ 4 = 2 \times 2 $
- $ 12 = 3 \times 4 $
- $ 30 = 5 \times 6 $
- $ 60 = 6 \times 10 $? Not clear.

Wait:
Check if it's related to $ n(n+1) $ or similar:

Try:
- $ a_1 = 4 = 2 \times 2 $
- $ a_2 = 12 = 3 \times 4 $
- $ a_3 = 30 = 5 \times 6 $
- $ a_4 = 60 = 6 \times 10 $? No.

Wait — let's try:
- $ a_1 = 4 = 1 \times 4 $
- $ a_2 = 12 = 2 \times 6 $
- $ a_3 = 30 = 3 \times 10 $
- $ a_4 = 60 = 4 \times 15 $

Now look at multipliers: 4, 6, 10, 15

These are:
- 4 = 1 + 3
- 6 = 2 + 4
- 10 = 3 + 7? Not helpful.

Alternatively, notice:
- $ a_n = n(n+1)(n+2)/something $? Try plugging in.

Actually, this sequence might be $ a_n = n(n+1)(n+2)/2 $?

Test:
- $ n=1: 1×2×3/2 = 3 $ → no
- $ n=2: 2×3×4/2 = 12 $ → matches $ a_2 $
- $ n=3: 3×4×5/2 = 30 $ → matches
- $ n=4: 4×5×6/2 = 60 $ → matches!

But $ a_1 = 4 $, but $ 1×2×3/2 = 3 $ ≠ 4 → so not matching first term.

So maybe offset?

Wait — perhaps it's $ a_n = (n+1)(n+2)(n+3)/2 $? Let's test:
- $ n=1: 2×3×4/2 = 12 $ → too big

Alternatively, suppose the sequence starts at $ n=0 $?

Try $ a_0 = 4 $? But given $ a_1 = 4 $

Wait — what if we define:
- $ a_1 = 4 $
- $ a_2 = 12 $
- $ a_3 = 30 $
- $ a_4 = 60 $

Let’s compute ratios again:
- $ a_2/a_1 = 3 $
- $ a_3/a_2 = 2.5 $
- $ a_4/a_3 = 2 $

Not geometric.

Differences:
- $ \Delta_1 = 8 $
- $ \Delta_2 = 18 $
- $ \Delta_3 = 30 $

Now second differences:
- $ 18 - 8 = 10 $
- $ 30 - 18 = 12 $

Third differences:
- $ 12 - 10 = 2 $

Constant third difference suggests cubic polynomial.

Assume $ a_n = an^3 + bn^2 + cn + d $

Use $ n=1,2,3,4 $:

For $ n=1 $: $ a + b + c + d = 4 $
$ n=2 $: $ 8a + 4b + 2c + d = 12 $
$ n=3 $: $ 27a + 9b + 3c + d = 30 $
$ n=4 $: $ 64a + 16b + 4c + d = 60 $

Solve system:

Eq1: $ a + b + c + d = 4 $
Eq2: $ 8a + 4b + 2c + d = 12 $
Eq3: $ 27a + 9b + 3c + d = 30 $
Eq4: $ 64a + 16b + 4c + d = 60 $

Subtract Eq1 from Eq2:
(8a+4b+2c+d) - (a+b+c+d) = 12 - 4 → $ 7a + 3b + c = 8 $ → Eq5

Eq3 - Eq2:
(27a+9b+3c+d) - (8a+4b+2c+d) = 30 - 12 → $ 19a + 5b + c = 18 $ → Eq6

Eq4 - Eq3:
(64a+16b+4c+d) - (27a+9b+3c+d) = 60 - 30 → $ 37a + 7b + c = 30 $ → Eq7

Now Eq6 - Eq5:
(19a+5b+c) - (7a+3b+c) = 18 - 8 → $ 12a + 2b = 10 $ → $ 6a + b = 5 $ → Eq8

Eq7 - Eq6:
(37a+7b+c) - (19a+5b+c) = 30 - 18 → $ 18a + 2b = 12 $ → $ 9a + b = 6 $ → Eq9

Eq9 - Eq8:
(9a + b) - (6a + b) = 6 - 5 → $ 3a = 1 $ → $ a = 1/3 $

Plug into Eq8: $ 6*(1/3) + b = 5 $ → $ 2 + b = 5 $ → $ b = 3 $

From Eq5: $ 7*(1/3) + 3*3 + c = 8 $ → $ 7/3 + 9 + c = 8 $ → $ c = 8 - 9 - 7/3 = -1 - 7/3 = -10/3 $

From Eq1: $ 1/3 + 3 - 10/3 + d = 4 $ → $ (1/3 - 10/3) + 3 + d = 4 $ → $ -9/3 + 3 + d = 4 $ → $ -3 + 3 + d = 4 $ → $ d = 4 $

So $ a_n = \frac{1}{3}n^3 + 3n^2 - \frac{10}{3}n + 4 $

This is messy — probably not intended.

Wait — reconsider:
Is it possible that the sequence is $ a_n = n(n+1)(n+2)/2 $? But earlier didn’t match.

Wait — try:
- $ a_1 = 1×2×3 = 6 $ → too big
- $ a_1 = 2×2 = 4 $
- $ a_2 = 3×4 = 12 $
- $ a_3 = 5×6 = 30 $
- $ a_4 = 6×10 = 60 $? No.

Wait:
$ a_1 = 4 = 2×2 $
$ a_2 = 12 = 3×4 $
$ a_3 = 30 = 5×6 $
$ a_4 = 60 = 6×10 $? No.

Wait — factor:
- 4 = 2×2
- 12 = 3×4
- 30 = 5×6
- 60 = 6×10? No.

Wait — 4, 12, 30, 60

Notice:
- $ a_1 = 4 $
- $ a_2 = 12 = 4 × 3 $
- $ a_3 = 30 = 12 × 2.5 $
- $ a_4 = 60 = 30 × 2 $

Not helpful.

Wait — look at:
- $ a_1 = 4 = 1×4 $
- $ a_2 = 12 = 2×6 $
- $ a_3 = 30 = 3×10 $
- $ a_4 = 60 = 4×15 $

Now: 4, 6, 10, 15

These are:
- 4 = 2×2
- 6 = 2×3
- 10 = 2×5
- 15 = 3×5

Or: 4, 6, 10, 15 — differences: 2, 4, 5 — not clear.

Wait — 4 = 1+3, 6 = 2+4, 10 = 3+7, 15 = 4+11 — no.

Alternatively, consider:

Let me search for known sequences.

Wait — perhaps it's $ a_n = n(n+1)(n+2)/2 $? For $ n=2 $: $ 2×3×4/2 = 12 $, $ n=3: 3×4×5/2 = 30 $, $ n=4: 4×5×6/2 = 60 $, $ n=1: 1×2×3/2 = 3 $ — but first term is 4, not 3.

So maybe the sequence starts at $ n=2 $? But it says $ a_1 = 4 $

Unless there's a typo.

Alternatively, could it be $ a_n = (n+1)(n+2)(n+3)/2 $? For $ n=1 $: $ 2×3×4/2 = 12 $ — too big.

Wait — what if it's $ a_n = n(n+1)(n+2)/2 $ starting from $ n=1 $: 3, 12, 30, 60 — so maybe the first term is missing? But problem says 4, 12, 30, 60...

So unless it's not that.

Wait — another idea:
Maybe the sequence is defined recursively as:
- $ a_1 = 4 $
- $ a_{n} = a_{n-1} + 2n(n+1) $? Try:
- $ a_2 = 4 + 2×2×3 = 4 + 12 = 16 $ → no

Alternatively, differences:
- $ a_2 - a_1 = 8 $
- $ a_3 - a_2 = 18 $
- $ a_4 - a_3 = 30 $

Now: 8, 18, 30

Differences: 10, 12 — increasing by 2 → next would be 14, then 16

So second differences increase by 2 → quadratic differences → cubic sequence.

We already found the cubic formula.

But perhaps it's better to accept that the closed form is messy, and focus on recursive.

But let’s see if there's a pattern.

Another idea:
Check if it's $ a_n = 2n(n+1) $:

- $ n=1: 2×1×2 = 4 $ ✔
- $ n=2: 2×2×3 = 12 $ ✔
- $ n=3: 2×3×4 = 24 $ ✘ (should be 30)

No.

Try $ a_n = n(n+1)(n+2)/2 $:
- $ n=1: 3 $
- $ n=2: 12 $
- $ n=3: 30 $
- $ n=4: 60 $

So if the sequence was 3, 12, 30, 60, then yes.

But it starts with 4.

So maybe it's $ a_n = \frac{n(n+1)(n+2)}{2} $ for $ n \geq 2 $, and $ a_1 = 4 $? Unlikely.

Alternatively, perhaps the first term is off.

Wait — maybe it's a typo, and it should be 3, 12, 30, 60?

But let’s assume it's correct: 4, 12, 30, 60

Let’s try to find a recursive pattern.

Suppose:
- $ a_1 = 4 $
- $ a_2 = 12 = 4 × 3 $
- $ a_3 = 30 = 12 × 2.5 $
- $ a_4 = 60 = 30 × 2 $

Ratios: 3, 2.5, 2 — decreasing by 0.5

So maybe $ a_n = a_{n-1} \times (3.5 - 0.5(n-1)) $?

For $ n=2 $: $ 3.5 - 0.5(1) = 3 $ → 4×3=12 ✔
$ n=3 $: $ 3.5 - 1 = 2.5 $ → 12×2.5=30 ✔
$ n=4 $: $ 3.5 - 1.5 = 2 $ → 30×2=60 ✔

So general: $ a_n = a_{n-1} \times (3.5 - 0.5(n-1)) = a_{n-1} \times (4 - 0.5n) $

But 3.5 - 0.5(n-1) = 3.5 - 0.5n + 0.5 = 4 - 0.5n

Yes.

So $ a_n = a_{n-1} \times (4 - 0.5n) $

But this is ugly.

Better to write as:

$ a_n = a_{n-1} \times \left( \frac{8 - n}{2} \right) $

Check:
- $ n=2 $: $ (8-2)/2 = 3 $ → 4×3=12 ✔
- $ n=3 $: $ (8-3)/2 = 2.5 $ → 12×2.5=30 ✔
- $ n=4 $: $ (8-4)/2 = 2 $ → 30×2=60 ✔

So recursive formula:
$$
a_1 = 4, \quad a_n = a_{n-1} \cdot \frac{8 - n}{2}, \quad n \geq 2
$$

But this is unusual.

Alternatively, since we have a cubic pattern, perhaps it's better to accept that.

But let's move on — perhaps this is not the intended path.

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Given the complexity, and since you mentioned "I uploaded an image", I suspect the actual problems are clearer.

But since I can't see the image, let me assume the following common types of problems based on the layout:

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Let me solve the remaining problems assuming they are standard.

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Problem 2:

Write a recursive formula for:
2, 24, 24×(24+1)/2 = 24×25/2 = 300, ...

Wait, likely typo.

"2, 24, 24×(24+1)/2"

That’s strange.

Wait — probably:
Sequence: 2, 24, 300, ...

But 24×(24+1)/2 = 24×25/2 = 300

So $ a_1 = 2 $, $ a_2 = 24 $, $ a_3 = 300 $

But how to get from 2 to 24? 2×12 = 24

Then 24 to 300? 24×12.5 = 300

Not nice.

Perhaps it's:
$ a_1 = 2 $
$ a_2 = 24 $
$ a_3 = 24 \times \frac{24+1}{2} = 24 \times 12.5 = 300 $

So maybe the rule is: $ a_{n} = a_{n-1} \times \frac{a_{n-1} + 1}{2} $?

Check:
- $ a_2 = a_1 \times (a_1 + 1)/2 = 2 × 3/2 = 3 $ → not 24

No.

Wait — perhaps it's $ a_n = a_{n-1} \times (a_{n-1}) $? 2→4, not 24.

Alternatively, maybe it's a factorial or exponential.

But let's skip and go to others.

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Problem 3:

Write a recursive formula for:
-2, 4, -8, 0, ...

Wait — -2, 4, -8, 0 — seems irregular.

-2 to 4: ×(-2)
4 to -8: ×(-2)
-8 to 0: +8? Not consistent.

Unless it's alternating sign and doubling, but last term is 0.

Maybe typo? Perhaps it's -2, 4, -8, 16 — then $ a_n = -2 \times a_{n-1} $

But given 0, so maybe not.

Perhaps it's: -2, 4, -8, 0 — but 0 breaks pattern.

Alternatively, maybe it's arithmetic after some point? Unlikely.

Skip.

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Problem 4:

Write a recursive formula for:
4, 9, 66, 7401, ...

This is wild.

4 to 9: +5
9 to 66: +57
66 to 7401: +7335 — huge jump.

Not helpful.

Perhaps it's a known sequence.

But likely a typo or misread.

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Problem 5:

Write a recursive formula for:
1, 21, 677, ...

1 to 21: ×21
21 to 677: ≈32.23 times

Not obvious.

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Problem 6:

Find the first 4 terms of the sequence:
$ a_1 = 5 $, $ a_n = 11 - 2a_{n-1} $

Compute:
- $ a_1 = 5 $
- $ a_2 = 11 - 2×5 = 11 - 10 = 1 $
- $ a_3 = 11 - 2×1 = 11 - 2 = 9 $
- $ a_4 = 11 - 2×9 = 11 - 18 = -7 $

So: 5, 1, 9, -7

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Problem 7:

$ a_1 = 3 $, $ a_n = (-1)^n \cdot a_{n-1} $

Compute:
- $ a_1 = 3 $
- $ a_2 = (-1)^2 × a_1 = 1 × 3 = 3 $
- $ a_3 = (-1)^3 × a_2 = -1 × 3 = -3 $
- $ a_4 = (-1)^4 × a_3 = 1 × (-3) = -3 $

So: 3, 3, -3, -3

Note: $ (-1)^n $, so for $ n=2 $: (+), $ n=3 $: (-), etc.

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Problem 8:

Find the first 4 terms of the sequence:
$ a_1 = 2 $, $ a_n = 2a_{n-1} + 3 $

Compute:
- $ a_1 = 2 $
- $ a_2 = 2×2 + 3 = 4 + 3 = 7 $
- $ a_3 = 2×7 + 3 = 14 + 3 = 17 $
- $ a_4 = 2×17 + 3 = 34 + 3 = 37 $

So: 2, 7, 17, 37

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Problem 9:

$ a_1 = 3 $, $ a_n = 12a_{n-1}/n $

Compute:
- $ a_1 = 3 $
- $ a_2 = 12×a_1 / 2 = 12×3 / 2 = 36 / 2 = 18 $
- $ a_3 = 12×a_2 / 3 = 12×18 / 3 = 216 / 3 = 72 $
- $ a_4 = 12×a_3 / 4 = 12×72 / 4 = 864 / 4 = 216 $

So: 3, 18, 72, 216

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Problem 10:

$ a_1 = 6 $, $ a_n = 10^n - a_{n-1} $

Compute:
- $ a_1 = 6 $
- $ a_2 = 10^2 - a_1 = 100 - 6 = 94 $
- $ a_3 = 10^3 - a_2 = 1000 - 94 = 906 $
- $ a_4 = 10^4 - a_3 = 10000 - 906 = 9094 $

So: 6, 94, 906, 9094

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Summary of Solvable Problems:

#### Problem 6:
Answer: $ 5, 1, 9, -7 $

#### Problem 7:
Answer: $ 3, 3, -3, -3 $

#### Problem 8:
Answer: $ 2, 7, 17, 37 $

#### Problem 9:
Answer: $ 3, 18, 72, 216 $

#### Problem 10:
Answer: $ 6, 94, 906, 9094 $

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For Problems 1–5:

Due to ambiguity in the image (e.g., unclear sequences, potential typos), I cannot confidently provide accurate recursive formulas without seeing the exact numbers.

However, here are general guidelines for writing recursive formulas:

- Arithmetic sequence: $ a_n = a_{n-1} + d $
- Geometric sequence: $ a_n = a_{n-1} \times r $
- Other patterns: Look for multiplicative or additive rules between terms.

If you can describe the sequences or type them out, I can help with problems 1–5.

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Would you like to re-upload the image or type the sequences clearly?