# Question 5 Exercise 5.3

Solutions of Question 5 of Exercise 5.3 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.

Find $n$ term and sum to $n$ terms each of the series $3+9+21+45+93+189+\ldots$

\begin{align} & a_2-a_1=9-3=6 \\ & a_3-a_2=21-9=12 \\ & a_4-a_3=45-21=24\\ & \text {... ... ... } \\ & \text {... ... ... } \\ &a_n-a_{n-1}=(\mathrm{n}-1)\quad \text{ term of the sequence}\quad 6,12,24, \ldots\end{align} which is a G.P. Adding column wise, we get \begin{align} a_n-a_1& =6+12+24+\ldots+(n-1) \text {terms } \\ & =\dfrac{6[2^{n-1}-1]}{2-1} \\ \Rightarrow a_n-a_1&=6 \cdot 2^{n-1}-6 \\ \Rightarrow a_n&=6 \cdot 2^{n-1}-6+a_1 \\ \Rightarrow a_n&=6 \cdot 2^{n-1}-6+3 \quad \because a_1=3\\ \Rightarrow a_n&=3(2^n-1)\end{align} Taking summation of the both sides \begin{align} \sum_{r=1}^n a_r&=6 \sum_{r=1}^n 2^{r-1}-3 \sum_{r=1}^n 1 \\ & =6 \cdot \dfrac{1 \cdot[2^n-1]}{2-1}-3 n \\ \Rightarrow \sum_{r=1}^n a_r&=6 \cdot(2^n-1)-3 n\\ \Rightarrow \sum_{r=1}^n a_r&=3(2^{n+1}-n-2)\\ \text{Hence}\quad 3(2^n-1);3(2^{n+1}-n-2)\end{align}