Answer to Question #147323 in Discrete Mathematics for Manzar abbas

Question

Use mathimatical Induction to prove that
1²+2²+3²+.......n²= n(n+1)(2n+1) divided by 6

Answer 1

First we need to show the result is true for smallest value of n, in this case n = 0/

When n =0 the left side has only one term n² = 0

The right side is

$\frac{n(n+2)(2n+1)}{6} = \frac{0(0+1)(0+1)}{6} = 0$

The second step is to show the statement is true for n = k, k+1 and so on.

Suppose, that statement is true for n = K.

1² + 2² + 3² + … + k² = $\frac{k(k+1)(2k+1)}{6}$

We need to prove, that statement is true fof n = k+1

1² + 2² + 3² + … + (k+1)² = $\frac{(k+1)(k+2)(2(k+1) +1)}{6}$

1² + 2² + 3² + … +k² + (k+1)² = $\frac{(k+1)(k+2)(2(k+1) +1)}{6}$

We know, that

1² + 2² + 3² + … +k² = $\frac{k(k+1)(2k+1)}{6}$

So,

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

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

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

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

$2k^2 + 7k + 6 = 0$

$2k^2 + 4k + 3k + 6 = 0$

$2k(k+2) + 3(k+2) =0$

$(2k+3)(k+2)=0$

$frac{k+1}{6}(2k^2 + 7k + 6) = \frac{(k+1)(k+2)(2k+3)}{6}$

$= \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$ = RHS hence proved