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:

Options

A. `(2n-1)/(2n)`
B. `(2n+1)/(2n-1)`
C. `(2n-1)/(2n+1)`
D. `(2n)/(2n+1)`

Answer

c

Explanation

Here ` S_(n)=u+a//2(2n-1)`
As u=0
`:. S_(n)=a//2(2n-1)`
Also `S_(n+1)=a//2 [2n(n+1)-1]=a//2[2n+1]`
Hence `(S_(n))/(S_(n+1))=(2n-1)/(2n+1)`