Question 8 Review Exercise

Solutions of Question 8 of Review Exercise of Unit 05: Miscullaneous Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find the sum of $n$ terms of the series whose $n^{t h}$ term is $n^3+3^n.$

The $n^h$ term is: $$a_n=n^3+3^n$$ Taking summation of the both sides $$\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+\sum_{r=1}^n 3^r \\ & =[\dfrac{n(n+1)}{2}]^2+\dfrac{3(3^n-1)}{3-1} \\ & =\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1) \end{align}$$ Thus the sum of $n$ terms is: $$S_n=\dfrac{n^2(n+1)^2}{4}+\dfrac{3}{2}(3^n-1)$$

Find the sum of $n$ terms of the series whose $n^{t h}$ term is $2 n^2+3 n$

The $n^{t h}$ term is: $$a_n=2 n^2+3 n$$ Taking summation on the both sides $$\begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^2+3 \sum_{r=1}^n r \\ & =2 \cdot[\dfrac{n(n+1)(2n+1)}{6}]++3 \dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)(2n+1)}{3}+\dfrac{3 n(n+1)}{2} \\ & =n(n+1)\dfrac{(2n+1)}{3} +\dfrac{3}{2} \\ & =n(n+1)\dfrac{2(2n+1)+9}{6}\\ & =\dfrac{n(n+1)(4n+11)}{6}\end{align}$$ Thus sum to $n$ terms is: $$S_n=\dfrac{n(n+1)(4n+11)}{6}$$

Find the sum of $n$ terms of the series whose $n^{t h}$ term is $n(n+1)(n+4)$

The $n^{\text {th }}$ term is: $$\begin{align} & a_n=n(n+1)(n+4) \\ & a_n==n(n^2+5 n+4) \\ & a_n=n^3+5 n^2+4 n\end{align}$$ Taking summation of the both sides $$\begin{align} \sum_{r=1}^n a_r&=\sum_{r=1}^n r^3+5 \sum_{r=1}^n r^2+4 \sum_{r=1}^n r \\ & =\dfrac{n^2(n+1)^2}{4}+5 \cdot \dfrac{n(n+1)(2 n+1)}{6}+4 \dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{n(n+1)}{2}+\dfrac{10 n+5}{3}+4] \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{3 n(n+1)+2(10 n+5)+24}{6}] \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{3 n^2+3 n+20 n+10+24}{6}] \\ & =\dfrac{n(n+1)(3 n^2+23 n+34)}{12}\end{align}$$ Thus the sum to $n$ terms is: $$S_n=\dfrac{n(n+1)(3 n^2+23 n+34)}{12}$$

Find the sum of $n$ terms of the series whose $n^{t h}$ term is $(2 n-1)^2$

The $n^{t h}$ term is: $$\begin{align} & a_n=(2 n-1)^2 \\ & a_n=4 n^2-4 n+1\end{align}$$ Taking summation of the both sides $$\begin{align} \sum_{r=1}^n a_r&=4 \sum_{r=1}^n r^2-4 \sum_{r=1}^n r+\sum_{r=1}^n 1 \\ & =4 \dfrac{n(n+1)(2 n+1)}{6}-4 \dfrac{n(n+1)}{2}+n \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{4(2 n+1)}{6}-2]+n \\ & =\dfrac{n(n+1)}{2} \cdot[\dfrac{8 n+4-12}{6}]+n \\ & =\dfrac{n(n+1)(8 n-8)}{12}+n \\ & =\dfrac{4n(n+1)(2 n-2)}{12}+n \\ & =\dfrac{4(n(2n^2-2n+2n-2))}{12}+n \\ & =\dfrac{n(2n^2-2)}{3}+n\\ & =\dfrac{n(2n^2-2+3)}{3}\\ & =\dfrac{n(2n^2+1)}{3}\end{align}$$ Thus the sum to $n$ terms is: $$S_n=\dfrac{n(2n^2+1)}{3}$$

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