Question

for all n>or =1 prove that 1^2+2^2+3^2+4^2+n^2=n (n+1)(2n+1)/6 please solve in copy and send
​
​

Answer 1

Step-by-step explanation:

Given, 1^2+2^2+3^2+4^2+n^2=n (n+1)(2n+1)/6

n = 1 => 1² = 1 (1+1) [2(1)+1] / 6

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

n = 2 => 1² + 2² = 2 (2+1) [2(2)+1] / 6

=> 1 + 4 = 2(3)(5) / 6

=> 5 = 30/6 => 5 = 5

n = 3 => 1² + 2² + 3² = 3 (3+1) [2(3)+1] / 6

=> 1 + 4 + 9 = 3(4)(7) / 6

=> 14 = 84/6 => 14 = 14

and continues

.•. for all n>or =1

1^2+2^2+3^2+4^2+n^2=n (n+1)(2n+1)/6