Question 3 Exercise 5.3

Solutions of Question 3 of Exercise 5.3 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 $n$ term and sum to $n$ terms each of the series $4+10+18+28+40+\ldots$

We use the method of difference as: $$\begin{align} & a_2-a_1=10-4=6 \\ & a_3-a_2=18-10=8 \\ & a_4-a_3=28-18=10 \\ & \text {... ... ... } \\ & \text {... ... ... } \\ & a_n-a_{n \quad 1}=(\mathrm{n}-1) \text { term of the sequence } \end{align}$$ $6,10,8, \ldots$ which is a A.P. Adding column wise, we get $$\begin{align}& a_n-a_1=6+10+8-\ldots +(n-1) \text { terms } \\ & =\dfrac{n-1}{2}[2 \cdot 6+(n-2) \cdot 2] \\ & =\dfrac{n-1}{2}[12+2 n-4] \\ & =\dfrac{n-1}{2}[2 n+8]=2 \cdot \dfrac{n-1}{2} \cdot[n+4] \\ & \Rightarrow a_n=n^2+3 n-4+a_1 \\ & \Rightarrow a_n=n^2+3 n-4+4 \quad \because a_1=4 \\ & \Rightarrow a_n=n^2+3 n\end{align}$$ Taking summation of the both sides $$\begin{align}\sum_{r=1}^n a_r&=\sum_{r=1}^n r^2+3 \sum_{r=1}^n r \\ & =\dfrac{n(n+1)(2 n+1)}{6}+3 \dfrac{n(n+1)}{2} \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1}{3}+3] \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+1+9}{3}] \\ & =\dfrac{n(n+1)}{2}[\dfrac{2 n+10}{3}] \\ & =\dfrac{n}{3}(n+1)(n+5)\\ \Rightarrow \sum_{r=1}^n a_r&=\dfrac{n}{3}(n+1)(n+5)\\ \text{Hence}\quad n^2+3 n;\dfrac{n}{3}(n+1)(n+5)\end{align}$$

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