Question

find the value of n in full process
$1 {}^{2} + 2 {}^{2} + 3 {}^{2} + 4 {}^{2}...........n {}^{2} = 1015$

Answer 1

Hey there!

Given,

1² + 2²+ ............+n² = 1015

$\sum \: {n}^{2} = 1015$
n ( n + 1 ) ( 2n+1 ) /6= 1015

[ A direct formula $\sum$ n² = n(n+1)(2n+1)/6 ]

Now,

n ( n + 1 ) ( 2n+ 1 ) /6= 5 * 203

n ( n+1 ) ( 2n+1) = 5 * 7 * 29*6

n ( n+1) ( 2n+1) = 5*7*2*3*29

n ( n+1) ( 2n+1) = (7*2)(5*3)(29)

n(n+1)(2n+1) = 14(15)(29)

n( n+1)(2n+1) = 14(14+1) ( 14*2 + 1 )

So, n = 14 .

If you know solving cubic equations, you can multiply n(n+1)(2n+1) and solve the equation. Mostly the above process is used. So that calculations don't go very far.

Hope helped!

Answer 2

Answer is given in the attachment