Question

solve it by using componendo and dividendo
$1 {}^{2} + 2 {}^{2} + 3 {}^{2} + .........n {}^{2} = 1015$
find n by using componendo and dividendo

Answer 1

We know that Sum of squares of first n natural numbers:

n(n + 1)(2n+1)/6  ------- (1)

Now,

$\frac{n(n+1)(2n+1)}{6} = 1015$

n(n+1)(2n+1) = 1015 * 6

n(n+1)(2n+1) = 6090

2n^3 + 3n^2 + n = 6090

2n^3 + 3n^2 + n - 6090 = 0

(n - 14)(2n^2 + 31n + 435) = 0

n = 14.


Hope this helps!