# Question 10 Review Exercise

Solutions of Question 10 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 10

Find the $n^{\text {th }}$ term and the sum to $n$ terms of the series $1+(1+\dfrac{1}{2})+(1+\dfrac{1}{2}+\dfrac{1}{4})+(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8})+\ldots$

### Solution

The general term of the given series is: \begin{align} a_n&=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots+\dfrac{1}{2^{n-1}} \\ a_n&=\dfrac{1[1-(\dfrac{1}{2})^n]}{1-\dfrac{1}{2}} \\ a_n&=2 \cdot[1-\dfrac{1}{2^n}] \\ \Rightarrow a_n&=2-\dfrac{1}{2^{n-1}}\\ \Rightarrow a_n&=2(1-\dfrac{1}{2^{n}})\end{align} Taking summation on the both sides \begin{align} \sum_{r=1}^n a_i&=2 \sum_{r=1}^n 1-\sum_{r=1}^n \dfrac{1}{2^{r-1}} \\ & =2 n-\dfrac{1[1-(\dfrac{1}{2})^n]}{1-\dfrac{1}{2}} \\ & =2 n-2[r \dfrac{1}{2^n}]\end{align} Thus the sum to $n$ terms is: $$S_n=\dfrac{1}{2^{n-1}}+2 (n-1)$$ $$\text{Hence}\quad 2(1-\dfrac{1}{2^{n}});\dfrac{1}{2^{n-1}}+2 (n-1)$$

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