InterviewSolution
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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)`A. `S_(40)=-820`B. `S_(2n) gt S_(2n+2)`C. `S_(51)=1326`D. `S_(2n +1) gt S_(2n-1)` |
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Answer» Correct Answer - A::B::C::D 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)] |
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