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I need to get the sum of the next series:

$\sum_n^m(n*(\sum_n^mn)) \Rightarrow [n,m]\in \Bbb{N}$

So i am not able to make it explicit. But i can't combine the $\sum_n^m = \frac{n(n+m)}{2}$ formula inside of the sum. I need to get an explicit formula for calculations.

So if someone knows a method to get it, or can help me to get the formula i would happy.

PS.: The formula need to give back

$\sum_2^5(n*(\sum_n^5n)) = 2*(2+3+4+5)+3*(3+4+5)+4*(4+5)+5^2 = 188$

Kovács Gergely
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1 Answers1

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\begin{align} \sum_{k=n}^m \left(k \sum_{p=k}^m p\right)&=\sum_{k=n}^m \left(k \frac{m(m+1)-(k-1)k}2\right)\\ &=\frac 12\sum_{k=n}^m (m(m+1)k+k^2-k^3)\\ &=\frac 12\sum_{k=1}^m (m(m+1)k+k^2-k^3)-\frac 12\sum_{k=1}^{n-1} (m(m+1)k+k^2-k^3)\\ &=\frac 12m(m+1)\frac{(m(m+1)-(n-1)n)}2\\&\,+\frac12\frac{(m(m+1)(2m+1)-(n-1)n(2n-1))}6\\&\,-\frac12\frac{(m^2(m+1)^2-(n-1)^2n^2)}4\\ \end{align}

Kay K.
  • 9,931
  • Hm, after some calculations the result formula do not gives the expected result on a small range of numbers. fox example from 2 to 5...as in my example...it mus be 125...but trough your formula i have got 95.....can you check it also? i've got that result 3 times... – Kovács Gergely Dec 30 '15 at 22:24