Question 7 Review Exercise

Solutions of Question 7 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 the series: $1.2^2+3.3^2+5.4^2+\ldots$ to $n$ terms.

The given series if the product of corresponding terms of the two series $1,3,5, \ldots,(2 n-1)$ and $2^2, 3^2, 4^2, \ldots,(n+1)^2$.

Therefore, the general lerm of the series is: $$\begin{align} & a_n=(2 n-1)(n+1)^2 \\ & a_n=(2 n-1)(n^2+2 n+1) \\ & a_n=2 n^3+3 n^2-1\end{align}$$ Taking summation of the both sides, we get $$\begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^3+\sum_{r=1}^n r^2-\sum_{r=1}^n 1 \\ & =2[\dfrac{n(n+1)}{2}]^2+\dfrac{n(n+1)(2 n+1)}{6}-n \\ & =\dfrac{n^2(n+1)^2}{2}+\dfrac{n(n+1)(2 n+1)}{6}-n \\ & =\dfrac{3 n^2(n+1)^2}{6}+\dfrac{n(n+1)(2 n+1)}{6}-n \\ & =\dfrac{n(n+1)}{6} \cdot[3 n(n+1)+(2 n+1)]-n \\ & =\dfrac{n(n+1)}{6} \cdot[3 n^2+3 n+2 n+1]-n\\ & =\dfrac{n(n+1)(3 n^2+5 n+1)}{6}-n \\ & =n \cdot \dfrac{3 n^3+8 n^2+6 n+1-6}{6}\\ S_n&=\dfrac{n(3 n^3+8 n^2+6 n-5)}{6}\end{align}$$

Find the sum the series: $3.1^2+5.2^2+7.3^2+\ldots$ to $n$ terms.

In the given series each term is the product of the corresponding terms of the two series: $3,5,7, \ldots,(2 n-1)$ and $1^2, 2^2, 3^2, \ldots, n^2$.

Therefore, the $n^{\text {th }}$ term of the given series is: $$a_n=n^2 \cdot(2 n+1)\quad\text{or}\quad a_n=2 n^3+n^2$$ Taking summation of the both sides $$\begin{align} \sum_{r=1}^n a_r&=2 \sum_{r=1}^n r^3+\sum_{r=1}^n r^2 \\ & =2[\dfrac{n(n+1)}{2}]^2+\dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{n^2(n+1)^2}{2}+\dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{3 n^2(n+1)^2}{6}+\dfrac{n(n+1)(2 n+1)}{6} \\ & =\dfrac{n(n+1)}{6} \cdot[3 n(n+1)+(2 n+1)] \\ & =\dfrac{n(n+1)}{6} \cdot[3 n^2+3 n+2 n+1] \\ & =\dfrac{n(n+1)(3 n^2+5 n+1)}{6}\end{align}$$ Sum of the $n$ terms of the series is: $$S_n=\dfrac{n(n+1)(3 n^2+5 n+1)}{6}$$

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