Question 23 and 24, Exercise 4.7

Solutions of Question 23 and 24 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 to $n$ terms of the series (arithmetico-geometric series): $$1+2 \times 2+3 \times 2^{2}+4 \times 2^{3}+\ldots.$$

Solution.

Given arithmetic-geometric series: $$1+2 \times 2+3 \times 2^{2}+4 \times 2^{3}+\ldots$$

It can be written as $$1\times 1+2 \times 2+3 \times 2^{2}+4 \times 2^{3}+\ldots$$

The numbers $1,2,3,4,\ldots$ are in A.P. with $a=1$ and $d=1$.

The numbers $1, 2, 2^2, 2^3, \ldots$ are in G.P. with first term as 1 and $r=\frac{2}{1}=2$.

The sume of first $n$ terms of the arithmetico-geometric series is given by

$$\begin{align*} S_{n}=\frac{a}{1-r}+d r \frac{\left(1-r^{n}\right)}{(1-r)^{2}}-\frac{(a+n d) r^{n}}{1-r} \end{align*}$$ Thus $$\begin{align*} S_{n}&=\frac{1}{1-2}+(1)(2) \frac{\left(1-2^n\right)}{(1-2)^2}-\frac{(1+n (1)) (2)^{n}}{1-2} \\ &=-1+2-2\cdot 2^n+2^n+n\cdot 2^n \\ &= n\cdot 2^n -2^n + 1 \\ &=2^n(n-1)+1. \end{align*}$$ This is the required sum.

Sum to $n$ terms of the series (arithmetico-geometric series): $$1+4 y+7 y^{2}+10 y^{3}+\ldots$$

Solution.

The given arithmetic-geometric series is: $$1 + 4y + 7y^2 + 10y^3 + \ldots$$

The numbers $1, 4, 7, 10, \ldots$ are in A.P. with $a = 1$ and $d = 3$.

The numbers $1, y, y^2, y^3, \ldots$ are in G.P. with first term $1$ and $r = y$.

The sum of the first $n$ terms of the arithmetico-geometric series is given by: $$S_n = \frac{a}{1 - r} + d r \frac{1 - r^n}{(1 - r)^2} - \frac{(a + n d) r^n}{1 - r}$$

This gives $$\begin{align*} S_n & = \frac{1}{1 - y} + 3 \cdot y \cdot \frac{1 - y^n}{(1 - y)^2} - \frac{(1 + n \cdot 3) y^n}{1 - y} \\ &= \frac{1}{1 - y} + \frac{3y(1 - y^n)}{(1 - y)^2} - \frac{(1 + 3n)y^n}{1 - y}\\ &= \frac{(1 - y)}{(1 - y)^2} + \frac{3y - 3y^{n+1}}{(1 - y)^2} - \frac{(y^n + 3ny^n)(1 - y)}{(1 - y)^2}\\ &= \frac{(1 - y)+(3y - 3y^{n+1})-(y^n + 3ny^n-y^{n+1} -3n y^{n+1})}{(1 - y)^2}\\ &= \frac{1 - y+3y - 3y^{n+1}-y^n - 3ny^n+y^{n+1} +3n y^{n+1}}{(1 - y)^2}\\ &= \frac{3n y^{n+1} - 2y^{n+1} - 3ny^n-y^n+2y+1 }{(1 - y)^2} \end{align*}$$

This is the required sum.

< Question 21 & 22 Question 25 & 26 >