Telescoping series help

Question

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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)

TealSeal Mar 28, 2021

edited by TealSeal Mar 28, 2021

Guest · Answer 1 · 2021-03-28T05:09:19

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