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

Step-by-step explanation:

Proof by Mathematical Induction:

This involves the following steps:

• Prove true for some value, say n = 1
• Assume the result is true for n = k
• Prove true for n = k + 1

As per your given question:

n = 1   ⇒ L.H.S. = 1² = 1

and

$R.H.S.=\frac{1}{6}.1.(1+1)(2+1)$

$=\frac{6}{6}$

$=1$

∴ Result is true for n = 1

Assume result is true for n = k

assume 1² + 2² + .....+ k² = $\frac{1}{6}k(k+1)(2k+1)$

Prove true for n = k + 1

∴ $1^{2}+2^{2}+3^{2}+......+k^{2}+(k + 1)^{2}=\frac{1}{6}.k.(k+1)(2k+1)+(k+1)^{2}$

$=\frac{k.(k+1)(2k+1)+6(k+1)^{2}}{6}$

$=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}$

$=\frac{(k+1)[2k^{2}+k+6k+6]}{6}$

$=\frac{(k+1)[2k^{2}+7k+6]}{6}$

$=\frac{(k+1)(2k^{2}+4k+3k+6)}{6}$

$=\frac{(k+1)[2k(k+2)+3(k+2)]}{6}$

$=\frac{(k+1)(k+2)(2k+3)}{6}$

$=\frac{1}{6}(k+1)[(k+1)+1][2(k+1)+1]$

∴ Result is true for n= k + 1

⇒ 1² + 2² + 3² + .....+ n² = $\frac{1}{6}.n.(n+1)(2n+1)$ is true for any values of n.

___________________________

          //Hope this will Helped You//

        //Please Mark it as Brainliest//