find the sum-

1/(1+1^2+1^4)+2/(2+2^2+2^4)+3/(3+3^2+3^4)…99/(99+99^2+99^4)

# Haryana ntse question no.128 in 2014 paper

**anwesha**#2

(1/1+1^2+1^4)+(2/2+2^2+2^4)+(3/3+3^2+3^4)…(99/99+99^2+99^4)

= (1 + 1 + 1 +… 99 times) + (1^2 + 2^2 + 3^2 + … + 99^2) + (1^4 + 2^4 + 3^4 +… + 99^4)

= S_{1} + S_{2} + S_{3}

S_{1} = 1 + 1 + 1 + … + 1 (99 times) = 99

For S_{2} and S_{3}, we will have to use Power Summations.

**Sum of first n squares** can be written as **(1/6)n(n+1)(2n+1)**. In this case n being 99:

S_{2} = 328350

**Sum of first n to the power 4** can be written as ** (1/5)n5 + (1/2)n4 + (1/3)n3 - (1/30)n**. In this case n being 99:

Substitute n as 99 and find S_{3}. Then add S_{1}, S_{2}, S_{3} to get final sum!