# Question 19 and 20, Exercise 4.7

Solutions of Question 19 and 20 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 19

Sum the series up to $n$ term: $1^{3}+3^{3}+5^{3}+$

** Solution. **

#### Rough Work

Take $1+3+5+\ldots$.This is A.P with kth term $a_k=1+(k-1)(2)=1+2k-2=2k-1$.

Now consider this to make kth term of given series by just taking square.

Consider $T_k$ represents the $k$th term of the sereies, then \begin{align*}T_k&=(2k-1) \\ &=9k^2-6k+1. \end{align*}

Taking summation, we have \begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (2k - 1)\\ & = 2 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 \\ & = 2\left( \frac{n(n+1)}{2} \right) - n \\ & = n(n+1)= n \\ & = n^2 \end{align*}

Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} = n^2.$

## Question 20

Sum the series up to $n$ term: $2+5+10+17+\ldots$ to $n$ terms.

** Solution. **

Given: \begin{align*}&2+5+10+17\ldots\\ & (1^2+1)+(2^2+1)+(3^2+1)+(4^2+1)+\ldots\\ &(k^2+1)\end{align*}

Consider $T_k$ represents the $k$th term of the sereies, then \begin{align*}T_k&=k^2+1 \end{align*}

Taking summation, we have \begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (k^2 + 1)\\ & = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 1 \\ & = \frac{n(n+1)(2n+1)}{6} + n \\ & = \frac{n}{6}\left((n+1)(2n+1) + 6\right) \\ & = \frac{n}{6}\left(2n^2+3n+1 + 6\right) \\ & = \frac{n}{6}\left(2n^2+3n+7\right) \end{align*}

Thus, the sum of the series is $\sum\limits_{k=1}^{n} T_{k} = \dfrac{n}{6}\left(2n^2+3n+7\right) $

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