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.

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

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*}

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

<btn type=“primary”>< Question 19 & 20</btn> <btn type=“success”>Question 23 & 24 ></btn>