====== 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.

=====Question 4=====
Find $n$ term and sum to $n$ terms each of the series $3+5+11+29+83+245+\ldots$
====Solution====
\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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