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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