The value of m when the equation hold for any natural n.
$1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{m} .$

Open in App

Solution

Let the sum be denoted by S; then $S = 1^{2} + 2^{2} + 3^{2} . . . . . . . . . . . . n^{2}$
we have, $n^{3} - {(n - 1)}^{3} = {3 n}^{2} - 3 n + 1$
and then changing n to (n-1) we get a repeating sequence and
Hence, by adding we get
$n^{3} = 3 (1^{2} + 2^{2} + 3^{2} . . . . . . . . . . . . n^{2}) - 3 (1 + 2 + 3 + 4 + 5 + . . . . . . . . . . + n) + n = 3 S - \frac{3 n (n + 1)}{2} + n o r 3 S = n^{3} + \frac{3 n (n + 1)}{2} - n = n (n + 1) (n - 1 + \frac{3}{2}) o r S = \frac{n (n + 1) (2 n + 1)}{6}$
$m = 6$

Suggest Corrections

Similar questions

1^{2} + 2^{2} + 3^{2} + \dots \dots \dots \dots \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

n \in N

n \geq 1

1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

n \geq 1,

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

n \in N

1^{2} + 2^{2} + 3^{2} + . . . . . . . n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

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

Join BYJU'S Learning Program