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# Telescoping series help

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Here's the question

Let $a_n = 4n^3 + 6n^2 + 4n + 1.$ Find$a_8 + a_9 + a_{10} + \dots + a_{23}.$

Thank you.

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Source: alcumus

(I do not own this problem, it would be helpful if you show the process of solving instead of just the answer so we can all learn)

Mar 28, 2021
edited by TealSeal  Mar 28, 2021

### 2+0 Answers

#1
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You can use the formula

$\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$

$\sum_{k = 1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$

Then $a_8 + a_9 + \dots + a_{23} = \boxed{326520}$.

Mar 28, 2021
#2
+64
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The answer isn't right, but thanks for trying

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TealSeal  Mar 28, 2021
edited by TealSeal  Mar 28, 2021