Prove that (n(top)sigma r=1(bottom))^2-(n-1(top)sigma r=1(bottom))^2 =n^3 general formula for both notation is r

Answer 1

(sum(r) for r=1..n)^2 - (sum(r) for r=1..n-1)^2 = n^3

well, we know that

n
∑ r = n(n+1)/2
r=1
so, square that and we have

n^2(n+1)^2/4

Sum to n-1 instead of n, and we have

(n-1)^2 n^2/4

Subtract to get

n^2/4 ((n+1)^2 - (n-1)^2)
= n^2/4 (4n)
= n^3