====== Question 21 and 22, Exercise 4.7 ======
Solutions of Question 21 and 22 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 21=====
Sum the series up to $n$ term: $1 \times 4+2 \times 7+3 \times 10+\cdots$

** Solution. **

<callout type="tip" title="Rough Work" icon="true">
Take $4+7+10+\ldots$\\
This is A.P with kth term $a_k=4+(k-1)(3)=4+3k-3=3k+1$. \\
Also kth term of $1+2+3+...$ is $k$.
So we have required kth term $k(3k+1)$.
</callout>


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

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

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




=====Question 22=====
Sum the series up to $n$ term: $1 \times 3 \times 5+3 \times 5 \times 7+5 \times 7 \times 9+\cdots$ to $n$ term.

** Solution. **

<callout type="tip" title="Rough Work" icon="true">
Take $1+3+5+\ldots$.\\
This is A.P with kth term $a_k=1+(k-1)(2)=1+2k-2=2k-1$. \\
Take $3+5+7+\ldots$.\\
This is A.P with kth term $a_k=3+(k-1)(2)=3+2k-2=2k+1$. \\
Take $5+7+9+\ldots$.\\
This is A.P with kth term $a_k=5+(k-1)(2)=5+2k-2=2k+3$. \\
Thus required kth term is $(2k-1)(2k+1)(2k+3)$.\\
</callout>


Consider $T_k$ represents the $k$th term of the sereies, then
\begin{align*}T_k&=(2k-1)(2k+1)(2k+3) \\
&=(4k^2-1)(2k+3)\\
&=8k^3+12k^2-2k-3\end{align*}

Taking summation, we have
\begin{align*}\sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} (8k^3+12k^2-2k-3)\\
& = 8 \sum_{k=1}^{n} k^3 +12 \sum_{k=1}^{n} k^2-2 \sum_{k=1}^{n} k -3 \sum_{k=1}^{n} 1  \\
& = 8\left(\frac{n(n+1)}{2}\right)^3+12\left(\frac{n(n+1)(2n+1)}{6}\right)-2\left( \frac{n(n+1)}{2} \right) - 3n \\
& = \left(n(n+1)\right)^3+2\left(n(n+1)(2n+1)\right)-\left( n(n+1) \right) - 3n \\
&=n[\left(n^2(n^3+3n^2+3n+1)\right)+2\left(2n^2+3n+1\right)-\left( (n+1) \right) - 3] \\
& = n[n^5+3n^4+3n^3+n^2+4n^2+6n+2-n-1 - 3] \\
& = n[n^5+3n^4+3n^3+5n^2+5n-2] \\
& = n(n+2)(n^4+n^3+n^2+3n-1) \end{align*}FIXME

Thus, the sum of the series is 
$\sum\limits_{k=1}^{n} T_{k} =n(n+2)(n^4+n^3+n^2+3n-1).$









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