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)`

If S_n=1^2+(1^2+2^2)+(1^2+2^2+3^2)+ upto n terms, then S_n=(n(n+1)^2(n+2))/(12) b. S_n=(n(n-1)^2(n+2))/(12) c. S_(22)=3276 d. S_(22)=23275

Find as indicated in the following case : S_(n)=S_(n-1)-1,(ngt2), S_(1)=S_(2)=2,S_(5) .

S_(n) = (1+2+3+....+n)/( n) then S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ..... + S_(n)^(2) =

If S_(n) is the sum of the first n terms of an A.P. then : (a) S_(3n)=3(S_(2n)-S_n) (b) S_(3n)=S_n+S_(2n) (c) S_(3n)=2(S_(2n)-S_(n) (d) none of these

Let sum of n , 2n , 3n , terms of an A.P are S_(1), S_(2), S_(3) respectively. Prove that S_(3) = 3 (S_(2) - S_(1)) .

S_(2n+1)-S_(2) must be equal to (A)((S_(1)-S_(2)))/(3)(1+(1)/(2^(2n-1)))

If in an A.P., S_(2n)=3.S_(n) then S_(3n) : S _(n) = (a)5 (b) 6 (c)7 (d)8

If S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!) , then S_(50)=

If S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!) , then S_(50)=

If S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!) , then S_(50)=