Question

(1+2+3+4+5+..........+n)*(1^2+2^2+3^2+......+n^2)/(1^3+2^3+3^3+4^3+5^3+.........+n^3)
NEED CORRECT ANS.​

Answer 1

Sum of n natural number 1+ 2+ ... + n = n(n+1)/2,

Sum of squared n natural number = n(n+1)(2n+1)/6

Sum of cube n natural number = (n ( n +1)/2)^2

So,

The answer is,

= n(n+1)/2 * n(n+1)(2n+1)/6 / (n ( n +1)/2)^2

= 1/2 * (2n+1)/6 / ((1/2)^2

= (2n+1)/48