Question

How prove that without induction 1²+2²+3²+....+n²=n(n+1)(2n+1)/6

Answer 1

Hii mate

Here's ur answer

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

case 1:let n=1

then :p(1)=1^2=1

=1(1+1)(2×1+1)/6

=2×3/6

=1

for n=1 ,p(1) is true

assume that p(k) is true for positive integer k i.e.

p(k)=1^2+2^2+.....+k^2=k(k+1)(2k+1)/6

p(k+1)=[1^2+2^2 +3^2+.... +k^2] +(k+1)^2

=k(k+1)(2k+1)/6+(k+1)^2

=k(k+1)(2k+1)+6(k+1)^2

= (k+1)/6[k(2k+1)+6(k+1)]

=(k+1)/6(2k^2+k+6k+6)

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

=(k+1)/6 [(k+2)+(2k+3)]

=[(k+1)(k+2)(2(k+1)+1]