Question

Derive a formula for

n

∑

i= i³ a telescoping sum with terms f(I) = i⁴

Answer 1

TO DERIVE :

let,

$S=\displaystyle\sum_{i=1}^ni^3$ ............(1)

We start off the proof with the "telescoping sum" (or the collapsing sum).

$\displaystyle\sum_{i=1}^n [(i+1)^4-i^4]$ SOLVING IN GENERAL FORM

$=(\cancel{2^4}-1^4)+(\cancel{3^4}-\cancel{2^4})+(4^4-\cancel{3^4})\\\\+...........+................+(n+1)^4-\cancel{n^4}\\+...........+................+ \cancel{(n)^4}-\cancel{(n-1)^4}\\ =(n+1)^4-1^4\\ =(n=1)^4-1\\ =((n+1)^2)^2-1\\ =(n^2+2n+1)^2-1$

$\displaystyle\sum_{i=1}^n [(i+1)^4-i^4]$ $=n^4+4n^3+6n^2+4n+1$ .......................(2)

now,

$\displaystyle\sum_{i=1}^n [(i+1)^4-i^4]$ SOLVING IN terms of i

$​ =\displaystyle\sum_{i=1}^n[(i^4+4i^3+6i^2+4i+1)-i^4]\\ =\displaystyle\sum_{i=1}^n [(4i^3+6i^2+4i+1)] \\ =4\displaystyle\sum_{i=1}^ni^3+6\displaystyle\sum_{i=1}^ni^2+4\displaystyle\sum_{i=1}^ni+\displaystyle\sum_{i=1}^n1\\$

from (1) ..further solving above...

$=4S+6({n(n+1)(2n+1)\over6})+4({n(n+1)\over2})+n$

$=4S+2n^3+n^2+2n^2+n+2n^2+3n\\ =4S+2n^3+5n^2+4n$ ................(3)

THUS; equating (2) and (3);

$n^4+4n^3+6n^2+4n=4S+2n^3+5n^2+4n\\ n^4+2n^3+n^2=4S\\ n^2(n^2+2n+1)=4S$

$4S=n^2(n+1)^2\\ S={n^2(n+1)^2\over2^2}$

$\boxed{S=[{n(n+1)\over2}]^2}$

hence drived.