Question

use the rule of mathematical induction to prove 1^2+2^2+3^2+....+n^2=n(n+1)(2n+1)/6​

Answer 1

Explanation:

For n = 1

LHS = 1² = 1

RHS = 1(1+1)(2*1 + 1)/6

= 1 * 2 * 3 / 6 = 1

Therefore, LHS = RHS

The result is true for n = 1

Let the result be true for n = k

Therefore, 1² + 2²+ 3² +....+ k² = k(k+1)(2k + 1)/6

Now, for n = k+1

LHS = (1² + 2² + 3² +.....+ k²) + (k + 1)2

= k(k+1)(2k + 1)/6 + (k+1)2 [From above result]

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

= (k+1) [ (2k2 + 7k + 6)/6 ]

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

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

Now, RHS = (k+1)(2k + 3)(k +2)/6

Therefore, LHS = RHS

So, The result is true for n = k+1

Therefore, by principle of mathematical induction, the result is true for n = any natural number. Proved.