If `S_n=1^2-2^2+3^2-4^2+5^2-6^2+ ,t h e n` `S_(40)=-820` b. `S_(2n)> S_(2n+2)` c. `S_(51)=1326` d. `S_(2n+1)> S_(2n-1)`

Options

A. `S_(40)=-820`
B. `S_(2n) gt S_(2n+2)`
C. `S_(51)=1326`
D. `S_(2n +1) gt S_(2n-1)`

Answer

A, B, C, D

Explanation

Clearly, nth term of the given series is negative or positive accordingly as n is even or odd, respectively.
Case I: When n is even: In this case, the given series is
`S_(n)=1^(2)-2^(2)+3^(2)-4^(2)+…+(n-1)^(2)-n^(2)`
`=(1^(2)-2^(2))+(3^(2)-4^(2))+..+((n-1)^(2)-n^(2))`
`=(1-2)(1+2)+(3-4)(3+4)+...+((n-1)-(n))(n-1+n)`
`=-(1+2+3+4+...+(n-1)+n)`
`=-(n(n+1))/2` (1)
Case II, When n is odd: In this case, the given series is
`S_(n)=(1^(2)-2^(2))+(3^(2)-4^(2))+..+{(n-2)^(2)-(n-1)^(2)}+n^(2)`
=`(1-2)(1+2)+(3-4)(3+4)+...+((n-2)-(n-1))xx((n-2)+(n-1))+n^(2)`
`=-((n-1)(n-1+1))/2+n^(2)=(n(n+1))/2` (2)
`rArrS_(40)=-820` [Using (1)]
`S_(51)=1326` [Using (2)]
Also, `S_(2n)gtS_(2n+2)` [From (1)]
`S_(2n+1)gtS_(2n+1)` [From (2)]