A small block slides without friction down an iclined plane starting form rest. Let `S_(n)` be the distance traveled from time `t = n - 1` to `t = n`. Then `(S_(n))/(S_(n + 1))` is:

A small block slides without friction down an inclined plane starting from rest. Lets s_n be the distance travelled from time t=n - 1 to t = n. Then (S_n)/(s_n+1) is:

A particle starts sliding down a frictionless inclined plane. If S_(n) is the distance travelled by it from time t = n-1 sec , to t = n sec , the ratio (S_(n))/(s_(n+1)) is

A particle starts sliding down a frictionless inclined plane. If S_(n) is the distance travelled by it from time t = n-1 sec , to t = n sec , the ratio (S_(n))/(s_(n+1)) is

A particle starts sliding down a frictionless inclined plane. If S_(n) is the distance travelled by it from time t = n-1 sec , to t = n sec , the ratio (S_(n))/(s_(n+1)) is

A particle starts sliding down a frictionless inclined plane. If S_(n) is the distance travelled by it from time t = n-1 sec , to t = n sec , the ratio (S_(n))/(s_(n+1)) is

A small cube falls from rest along a frictionless inclined plane. If this distance travelled between times t = n -1 and t = n be s_(n) then the value of (s_(n))/(s_(n)+1) is

In an A.P. if a=2, t_(n)=34 ,S_(n) =90 , then n=

In an A.P. if a=2, t_(n)=34 ,S_(n) =90 , then n=

If sum of n termsof a sequende is S_n then its nth term t_n=S_n-S_(n-1) . This relation is vale for all ngt-1 provided S_0= 0. But if S_!=0 , then the relation is valid ony for nge2 and in hat cast t_1 can be obtained by the relation t_1=S_1. Also if nth term of a sequence t_1=S_n-S_(n-1) then sum of n term of the sequence can be obtained by putting n=1,2,3,.n and adding them. Thus sum_(n=1)^n t_n=S_n-S_0. if S_0=0, then sum_(n=1)^n t_n=S_n. On the basis of above information answer thefollowing questions:If nth term of a sequence is n/(1+n^2+n^4) then the sum of its first n terms is (A) (n^2+n)/(1+n+n^2) (B) (n^2-n)/(1+n+n^2) (C) (n^2+n)/(1-n+n^2) (D) (n^2+n)/(2(1+n+n^2)