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How to calculate the sum for incremental value that is calculated by this formula:

$val = (5n)^2$ so sequence of $val$ will be $(25, 100, 225, 400, 625, \cdots)$

And sum sequence should be: $(25, 125, 350, 750, 1375, \cdots)$

(where sum is the total from 1 to $n$)

What is the formula to get the sum for $n$ when $n$ is $10$ or $15$?

jimjim
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xYuri
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  • Are you familiar with formulas for $\sum\limits_n n$ and $\sum\limits_n n^2$? – zkutch Jun 12 '21 at 00:12
  • @zkutch sorry no, I have simplified the formula for better understanding – xYuri Jun 12 '21 at 00:14
  • You can find it here https://math.stackexchange.com/questions/48080/sum-of-first-n-squares-equals-fracnn12n16 or here https://proofwiki.org/wiki/Sum_of_Sequence_of_Squares – zkutch Jun 12 '21 at 00:19
  • @zkutch I don't really understand how do I apply this for what I need, could you please apply it to my equation so I get better understanding? – xYuri Jun 12 '21 at 00:56
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    $\sum\limits_{k=1}^{n}(k\cdot 5)^2=5^2\sum\limits_{k=1}^{n}k^2$ – zkutch Jun 12 '21 at 01:08

1 Answers1

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You're looking for the sum of $(5n)^2 = 25n^2$, where $n$ ranges between 1 and some $x$.

So we can write this in sigma notation as $\sum_{n=1}^{x} 25n^2$, and we can pull out the constant factor of 25 to get $25\sum_{n=1}^{x}n^2$.

Using the well-known sum of squares formula $\sum_{n=1}^x n^2 = x(x+1)(2x+1)/6$, you have $25\sum_{n=1}^{x}n^2 = \boxed{\frac{25}{6}x(x+1)(2x+1)}$.

recursivecorgi
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  • Thanks, pretty good explanation, specially mentioning sigma notation after searching that, I get it now – xYuri Jun 12 '21 at 01:48