Question

Using the principles of mathematical induction prove:
1^2+2^2+3^2....n terms = n(n+1)(2n+1)/6

Answer 1

1^2+2^2+3^2+....+n^2 
We know that 
(x+1)^3-x ^3= 3x^2+3x+1 
Putting x=1,2.....n, we get 
2^3-1^3=3(1)^2+3(1)+1 
3^3-2^3=3(2)^2+3(2)+1 
.............................. 
............................ 
(n+1)^3-n^3=3(n)^2+3(n)+1 

Adding all the above terms, We get 
(n+1)^3 -1=3(1^2+2^2+3^2+........n^2)+3(1+2+3+..... 
or n^3+3n^2+3n =3(1^2+2^2+3^2+........n^2)+3(1+2+3+....... 
=3(1^2+.....+n^2)+3n(n+1)/2 +n 
or n^3+3n(n+1)= 3(1^2+.....+n^2)+3n(n+1)/2 +n 
or 3(1^2+.....+n^2)= n^3+3n(n+1)−3n(n+1)/2 −n 
=n^3+3n(n+1)/2 −n 
= n[n²+3(n+1)/2 −1]=n[2n²+3n+3 −2]/2 
=n[2n²+3n+1]/2 =n[2n²+2n+n +1]/2 
=n[2n(n+1)+(n +1)]/2=n(2n+1)(n+1)/2 
and (1^2+.....+n^2)=n(2n+1)(n+1)/6 =n(n+1)(2n+1)/6