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sum(k=1)^(2n+1)(-1)^(k-1)k^2=...

`sum_(k=1)^(2n+1)(-1)^(k-1)k^2=`

A

`(n-1)(2n-1)`

B

`(n+1)(2n+1)`

C

`(n+1)(2n-1)`

D

`(n-1)(2n+1)`

Text Solution

Verified by Experts

The correct Answer is:
b
` sum _(k=1)^(2n+1) (-1)^(k-1)k^(2)=[1^(2)-2^(2) +3^(2)-4^(2)+...-(2n)^(2)+(2n+1)^(2)]`
`= [ 1^(2) +3^(2)+5^(2) +...+ (2n+1)^(2) ] - [ 2^(2) +4^(2) +6^(2)+...+(2n)^(2)]`
`= [ 1^(2) +2^(2) +3^(2) +4^(2) +...+ (2n+1)^(2)]`
` -2[ 2^(2) +4^(2)+6^(2)+... + (2n)^(2)]`
` sum (2n+1)^(2) -2 xx 2^(2)[ 1^(2)+2^(2)+3^(2)+...+n^(2)]`
` = ((2n+1)(2n+1+1){2(2n+1)+})/6 [:' sum (n (n+1)(2n+1))/6]`
` = ((2n+1)(2n+2)(4n+3))/6 -8 [ (n(n+1)(2n +1))/6]`
` = ((n+1)(2n+1) )/6 [ 2(4n+3)-8n]`
` = ((n+1)(2n+1))/6 [ 8n + 6- 8n ] = (n+1) (2n+1)`
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