Suppose a series of `n` terms given by `S_(n)=t_(1)+t_(2)+t_(3)+`. . ..`+t_(n)`
then `S_(n-1)=t_(1)+t_(2)+t_(3)+` . . . `+t_(n-1),nge1` subtracting we get `S_(n)-S_(n-1)=t_(n),nge2` surther if we put `n=1` is the first sum then `S_(1)=t_(1)` thus w can write `t_(n)=S_(n)-S_(n-1),nge2` and `t_(1)=S_(1)`
Q. if the sum of `n` terms of a series `a.2^(n)-b` then the sum `sum_(r=1)^(infty)(1)/(t_(r))` is

Options

A. `a`
B. `(a)/(2)`
C. `(2)/(a)`
D. `(1)/(a)`

Answer

C

Explanation

`t_(1)=S_(1)=2a-b`
`t_(n)=S_(n)-S_(n-1)=a.2^(n)-a.2^(n-1)=a.2^(n-1),ngt1`
Now `(t_(n+1))/(t_(n))=2,nge2`
`sum_(r=2)^(infty)(1)/(t_(r))=(1)/(a)sum_(r=2)^(infty)(1)/(2^(r-1))=(1)/(a)(1//2)/(1-1//2)=(1)/(a)`