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)`

S_(n+3)-3S_(n+2)+3S_(n+1)-S_(n)=0

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

If S_(n)=1+3+6+10+...+(n(n+1))/(2) then S_(n) is

If S_(n)=1+3+6+10+...+(n(n+1))/(2) then S_(n) is

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_(3n)=2S_n then find the value of S_(4n)/S_(2n)

S_(n)=sum_(n=1)^(n)(n)/(1+n^(2)+n^(4)) then S_(10)S_(20)

The 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))