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The sum to 50 terms of the series 3/1...

The sum to 50 terms of the series
`3/1^2+5/(1^2+2^2)+7/(1^+2^2+3^2)+….+… is `

A

`(k+1)/1`

B

`k/(k+1)`

C

`(6k)/(k+1)`

D

`(6(k-1))/k`

Text Solution

Verified by Experts

The correct Answer is:
c
` S_(n) = sum_(k=1)^(n) ((2k+1))/(1^(2)+2^(2)+3^(2)+...+k^(2))= sum_(k=1)^(n) ((2k+1)xx6)/(k(k+1)(2k+1))`
` = sum_(k =1)^(n) 6/(k(k+1))=6 sum_(k=1)^(n)[ 1/k-1/(k+1)]=6[1-1/(k+1)]`
` = 6* k/(k+1) = (6k)/(k+1)`
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