`1^2+2^2+3^2++n^2=(n(n+1)(2n+1))/6`

For all n gt= 1 , prove that, 1^2 + 2^2 + 3^2 + 4^2 + …….+n^2 = (n(n+1)(2n+1))/(6)

Prove the following by induction. 1^2 + 2^2 + ….+ n^2 = n(n+1)(2n+1)//6

Prove by the principle of mathematical induction that for all n in N : 1^2+2^2+3^2++n^2=1/6n(n+1)(2n+1)

The sum of the squares of the first n positive integers 1^2 + 2^2 + 3^2 + ………+ n^2 is (n(n +1)(2n+1))/(6) . What is the sum of the squares of the first 9 positive integers?

1 ^(2) + 2^(2) + 3^(2) + . . . + n^(2) = (n (n + 1) (2 n + 1))/( 6)

Statement-1: 1^(2)+2^(2)+....+n^(2)=(n(n+1)(2n+1))/(6)"for all "n in N Statement-2: 1+2+3....+n=(n(n+1))/(2),"for all"n in N

Statement-1: 1^(2)+2^(2)+....+n^(2)=(n(n+1)(2n+1))/(6)"for all "n in N Statement-2: 1+2+3....+n=(n(n+1))/(2),"for all"n in N

For all n geq1 , prove that 1^2+2^2+3^2+4^2+dotdotdot+n^2= (n(n+1)(2n+1))/6

1.3 + 2.4 + 3.5 + …… + n(n+2) =( n(n+1)(2n+7))/6

For all ,prove that ,prove that 1^(2)+2^(2)+3^(2)+4^(2)+...+n^(2)=(n(n+1)(2n+1))/(6)