Question 4 Exercise 5.3

Solutions of Question 4 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 $3+5+11+29+83+245+\ldots$

\begin{align} & a_2-a_1=5-3=2 \\ & a_3-a_2=11-5=6 \\ & a_4-a_3=29-11=18 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n-1}=(\mathrm{n}-1) \text { term ofthe sequence }\end{align} $6,10,18, \ldots$ which is a G.P. Adding column wise, we get \begin{align} & a_n-a_1=2+6+18+\ldots+(n-1) \text { terms } \\ & =\dfrac{2 \cdot[3^{n- 1}-1]}{3-1} \\ \Rightarrow a_n-a_1&=3^{n-1}-1 \\ \Rightarrow a_n&=3^{n-1}-1+a_{1} \\ & \Rightarrow a_n=3^{n-1}-1+3=3^{n-1}+2 \end{align} Taking summation of the both sides \begin{align} & \sum_{r=1}^n a_r=\sum_{r=1}^n 3^{r-1}+2 \sum_{r=1}^n 1 \\ & =\dfrac{1 \cdot[3^n-1]}{3-1}+2 n \\ & \Rightarrow \sum_{r=1}^n a_r=\dfrac{1}{2}(3^n-1)+2 n \\ \text{Hence}\quad 3^{n-1}+2; \dfrac{1}{2}(3^n-1)+2 n\end{align}

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