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Mathematics Question on Sum of First n Terms of an AP

The sum of the series 1+43+109+2827+....1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+.... upto nn terms is

A

76n+16−23.2n−1\frac{7}{6}n+\frac{1}{6}-\frac{2}{3.2^{n-1}}

B

53n+76−12.3n−1\frac{5}{3}n+\frac{7}{6}-\frac{1}{2.3^{n-1}}

C

n+12−12.3n−1n+\frac{1}{2}-\frac{1}{2.3^{n-1}}

D

n−12−13.2n−1n-\frac{1}{2}-\frac{1}{3.2^{n-1}}

Answer

n+12−12.3n−1n+\frac{1}{2}-\frac{1}{2.3^{n-1}}

Explanation

Solution

Given series is 1+43+109+2827+.... n1+\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+....\,n terms =1+(1+13)+(1+19)+(1+127)+.... n=1+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{9}\right)+\left(1+\frac{1}{27}\right)+....\,n terms =(1+1+1+.... +n terms)=\left(1+1+1+....\,+n\, terms\right) +(13+19+127+.... n terms)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....\,n \,terms\right) =n+13(1−13n)1−13=n+13×32[1−3−n]=n+\frac{\frac{1}{3}\left(1-\frac{1}{3^{n}}\right)}{1-\frac{1}{3}}=n+\frac{1}{3}\times\frac{3}{2}\left[1-3^{-n}\right] =n+12[1−3−n]=n+12−12.3n=n+\frac{1}{2}\left[1-3^{-n}\right]=n+\frac{1}{2}-\frac{1}{2.3^{n}}