Let `a=min{x^(2)+2x+3,x epsilonR}` and `b=lim_(x theta to 0)(1-cos theta)/(theta^(2)0`. The value fo `sum_(r=0)^(n)a^(r).b^(n-r)` is

Options

A. `(2^(n+1)-1)/(3xx2^n)`
B. `(2^(n+1)+1)/(3xx2^n)`
C. `(4^(n+1)-1)/(3xx2^n)`
D. none of these

Answer

C

Explanation

We have `x^2+2x+3=(x+1)^2+2ge 2` for all `x in R`.
`therefore a=2`.
`b=lim_(thetato0) (1-cos theta)/(theta^2)=(1)/(2)`
`therefore sum_(xto0)^(n) a^rb^(n-r)=sum_(r=0)^(n)2^r((1)/(2))^(n-r)=sum_(r=0)^(n)((1)/(2))^(n-2r)`
`rArr sum_(r=0)^(n)a^rb^(n-r)=((1)/(2))^n+((1)/(2))^(n-2)+((1)/(2))^(n-4)+.....+((1)/(2))^(n-2n)`
`rArr sum_(r=0)^(n)a^rb^(n-r)=((1)/(2))^n{(1-(2^2)^(n+1))/(1-4)}=((1)/(2))^n((4^(n+1)-1)/3)`
`rArrsum_(r=0)^(n)a^rb^(n-r) (4^(n+1)-1)/(3xx2^n)`