Haryana ntse question no.128 in 2014 paper

ntse

#1

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)


#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)
= S1 + S2 + S3

S1 = 1 + 1 + 1 + … + 1 (99 times) = 99

For S2 and S3, 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:

S2 = 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 S3. Then add S1, S2, S3 to get final sum!


#3

Pl. answer this question…I have reviwed that question


#4

find the sum 1/(3^2-1)+1/(5^2-1)+1(7^2-1)…1/(35^2-1)