====== 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. **

<callout type="tip" title="Rough Work" icon="true">
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.
</callout>


Consider $T_k$ represents the $k$th term of the sereies, then
\begin{align*}T_k&=(2k-1)^3 \\
&=(2k)^3+3(2k)^2(-1)+3(2k)(-1)^2+(-1)^3 \\
&=8k^3-12k^2+6k+1
\end{align*}

Taking summation, we have FIXME
\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (8k^3-12k^2+6k+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 
$\sum\limits_{k=1}^{n} T_{k} = n^2.$ FIXME

=====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\\\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+2n+n+1 + 6\right) \\
& = \frac{n}{6}\left(2n^2+3n+7\right) \end{align*}

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








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<text align="left"><btn type="primary">[[math-11-nbf:sol:unit04:ex4-7-p8|< Question 17 & 18]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit04:ex4-7-p11|Question 21 & 22 >]]</btn></text>