# Question 14, 15 and 16, Exercise 4.7

Solutions of Question 14, 15 and 16 of Exercise 4.7 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

## Question 14

Find the sum to $n$ terms of the series whose $n$th term is $n+1$.

** Solution. **

Consider $T_n$ represents the $n$th term of series, then $$ T_{n} = n+1. $$

Taking summation

\begin{align*}\sum_{n=1}^{\infty} T_{n} &= \sum_{n=1}^{\infty} (n+1)\\ & = \sum_{n=1}^{\infty} n + \sum_{n=1}^{\infty} 1 \\ & = \frac{n(n+1)}{2} + n \\ & = \frac{n(n+1)}{2} + \frac{2n}{2} \\ & = \frac{n^2 + 3n}{2} \\ \sum T_{n} & = \dfrac{n(n+3)}{2} \end{align*} Thus, the sum of the series is $\sum_{n=1}^{\infty} T_{n}= \dfrac{n(n+3)}{2}$.

## Question 15

Find the sum to $n$ terms of the series whose $n$th term is $n^{2}+2 n$.

** Solution. **

Consider $T_k$ represents the $k$th term of series, then
$$ T_{k} = k^{2} + 2k $$
Taking summation, we have

\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (k^{2} + 2k)\\ & = \sum_{k=1}^{n} k^{2} + 2 \sum_{k=1}^{n} k \\ & = \frac{n(n+1)(2n+1)}{6} + 2\left( \frac{n(n+1)}{2} \right) \\ & = \frac{n}{6}\left[(n+1)(2n+1)+6(n+1)\right] \\ & = \frac{n}{6}\left(2n^2+2n+n+1+6n+6\right) \\ & = \frac{n}{6}\left(2n^2+9n+7\right) \end{align*} Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} = \dfrac{n}{6}(2n^2+9n+7)$.

## Question 16

Find the sum to $\boldsymbol{n}$ terms of the series whose $\boldsymbol{n}$ th term is given: $3 n^{2}+2 n+1$

** Solution. **

Consider $T_k$ represents the $k$th term of series, then
$$ T_{k} = 3k^{2} + 2k + 1 $$
Taking summation, we have

\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (3k^{2} + 2k + 1)\\ & = 3\sum_{k=1}^{n} k^{2} + 2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \\ & = 3\left( \frac{n(n+1)(2n+1)}{6} \right) + 2\left( \frac{n(n+1)}{2} \right) + n \\ & = \frac{n(n+1)(2n+1)}{2} + n(n+1) + n \\ & = \frac{n}{2}\left[(n+1)(2n+1)+2(n+1) + 2\right]\\ & = \frac{n}{2}\left(2n^2+2n+n+1+2n+2+2\right) \\ & = \frac{n}{2}\left( 2n^2+5n+5 \right) \end{align*}

Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} = \frac{n}{2}\left(2n^2+5n+5\right).$

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