====== Question 9 Review Exercise ======
Solutions of Question 9 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 9(i)=====
Find the sum of the first $n$ terms of the series $3+7+13+21+31+\ldots$
====Solution====
Using method of differences to compute the sum of the given series.
\begin{align}
& a_2-a_1=7-3=4 \\
& a_3-a_2=13-7=6 \\
& a_4-a_3=21-13=8 \\
& \ldots \quad \ldots \quad \ldots \\
& \ldots \quad \cdots \quad \ldots \\
& a_n-a_{n-1}=(n-1) \text { term of the series } \\
& 4,6,8, \ldots
\end{align}
Adding column wise, we get
\begin{align}
a_n-a_1&=4+6+8+\cdots+2 n \\
& =\dfrac{n-1}{2}[2 \cdot 4+2 \cdot(n-2)] \\
& =\dfrac{n-1}{2}[8+2 n-4] \\
& =\dfrac{(n-1)(2 n+4)}{2} \\
& =(n-1)(n+2) \\
\Rightarrow a_n&=n^2+n-2+a_1 \\
\Rightarrow a_n&=n^2+n-2+3 \because a_1=3 \\
\Rightarrow a_n&=n^2+n+1\end{align}
Taking summation of the both sides
\begin{align}
\sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+\sum_{r=1}^n r+\sum_{r=1}^n 1 \\
& =\dfrac{n(n+1)(2 n+1)}{6}+\dfrac{n(n+1)}{2}+n \\
& =n \cdot[\dfrac{2 n^2+3 n+1+3(n+1)+6}{6}] \\
& =n \cdot[\dfrac{2 n^2+3 n+3 n+3+6}{6}] \\
& =n \cdot \dfrac{2 n^2+6 n+9}{6}\end{align}
Thus the sum to $n$ terms is:
$$S_n=\dfrac{n(2 n^2+6 n+9)}{6}$$

=====Question 9(ii)=====
Find the sum of the first $n$ terms of the series $2+5+14+41+\ldots$
====Solution====
Using method of differences to compute the sum of the given series.
\begin{align}
& a_2-a_1=5-2=3 \\
& a_3-a_2=14-5=9 \\
& a_4-a_3=41-14=27\end{align}
$a_n-a_{n-1}=(n-1)$ term of the series $3,9,27, \ldots$

Adding column wise, we get
$$a_n-a_1=3+9+27+\cdots+3^n$$
Which is a geometric series with common ratio $r=3$. Therefore,
\begin{align}
a_n-a_1&=\dfrac{3(3^{n-1}-1)}{3-1} \\
& =\dfrac{3^n-3}{2}=\dfrac{1}{2} 3^n-\dfrac{3}{2} \\
\Rightarrow a_n&=\dfrac{1}{2} 3^n-\dfrac{3}{2}+3 \because a_1=3 \\
\Rightarrow a_n&=\dfrac{1}{2} 3^n+\dfrac{3}{2}\end{align}
Taking summation on the both sides
\begin{align}
\sum_{r=1}^n a_r&=\dfrac{1}{2} \sum_{r=1}^n 3^r+\dfrac{3}{2} \sum_{r=1}^n 1 \\
& =\dfrac{1}{2} \cdot \dfrac{3[3^n-1]}{3-1}+\dfrac{3 n}{2} \\
& =\dfrac{3^{n+1}-3}{4}+\dfrac{3 n}{2}\\
& =\dfrac{3}{4}(3^n-1)+\dfrac{3 n}{2}\end{align}
Thus the sum to $n$ terms is:
$$S_n=\dfrac{3}{4}(3^n-1)+\dfrac{3 n}{2}$$



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