Please help with the following series math.

Question

Prove the sum of the series in terms of x:

1²+2²+3²+...........+x² = [(x(x+1)(2x+1)]/6.

Answer 1

Use proof by mathematical induction

Show it works for x=1

1^2 = (1)(2)3)/6 = 1

show it works for x=2

1^2 + 2^2 = (2)(3)(5)/6 = 5

Assume it works for 1^2+ ...+x^2

then need to show it works for 1^2+...+x^2 + (x+1)^2

1^2 + 2^2 +....x^2 + (x+1)^2 = x(x+1)(2x+1)/6 + (x+1)^2

rewrite the right side to (x+1)(x+2)(2x+3)/6

factor out (x+1) after rewriting (x+1)^2 as 6(x+1)^2/6 and put it all over a common denominator 6

(x+1)[x(2x+1) + 6(x+1)]/6

= (x+1)[2x^2 +x + 6x + 6]/6

= (x+1)(2x^2 + 7x + 6)/6

=(x+1)(x+2)(2x+3)/6

QED

Answer 2