Question 25 and 26, Exercise 4.7

Solutions of Question 25 and 26 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 $n$ term of the series (arithmetico-geometric series): $1+\frac{4}{7}+\frac{7}{7^{2}}+\frac{10}{7^{3}}+\ldots$

Solution.

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

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

The numbers $1, \frac{1}{7}, \frac{1}{7^2}, \frac{1}{7^3}, \ldots$ are in GP with first term $1$ and $r = \frac{1}{7}$.

The sum of the first $n$ terms of the arithmetico-geometric series is given by: $$\begin{align*} S_n& = \frac{a}{1 - r} + d \cdot r \cdot \frac{1 - r^n}{(1 - r)^2} - \frac{(a + n \cdot d) r^n}{1 - r}\\ &= \frac{1}{1 - \frac{1}{7}} + 3 \cdot \frac{1}{7} \cdot \frac{1 - \left(\frac{1}{7}\right)^n}{\left(1 - \frac{1}{7}\right)^2} - \frac{(1 + n \cdot 3) \left(\frac{1}{7}\right)^n}{1 - \frac{1}{7}}\\ &= \frac{7}{6} + \frac{21(1 - \frac{1}{7^n})}{36} - \frac{7(1 + 3n)}{6 \cdot 7^n}\\ &=\frac{7}{6} + \frac{7(7^n - 1)}{12 \times7^n } - \frac{1 + 3n}{6 \times 7^n}\\ &=\frac{2\times 7^{n+1}+7^{n+1}-7-2-6n}{12\times 7^n} \\ &=\frac{3 \times7^{n+1}-9-6n}{12\times 7^n}\\ &=\frac{3(7^{n+1}-3-2n)}{12 \times7^n}\\ &=\frac{7^{n+1}-3-2n}{4\times 7^n}\\ \end{align*}$$ This is required sum.

Sum to $n$ term of the series (arithmetico-geometric series): $1+\frac{7}{2}+\frac{13}{4}+\frac{19}{8}+\frac{25}{16}+\ldots$

Solution.

The given arithmetic-geometric series is: $$1 + \frac{7}{2} + \frac{13}{4} + \frac{19}{8} + \frac{25}{16} + \ldots$$

The numbers $1, 7, 13, 19, 25, \ldots$ are in AP with $a = 1$ and $d = 6$.

The numbers $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots$ are in GP with first term $1$ and $r = \frac{1}{2}$

The sum of the first $n$ terms of the arithmetico-geometric series is given by: $$\begin{align*} S_n &= \frac{a}{1 - r} + d \cdot r \cdot \frac{1 - r^n}{(1 - r)^2} - \frac{(a + n \cdot d) r^n}{1 - r}\\ &= \frac{1}{1 - \frac{1}{2}} + 6 \cdot \frac{1}{2} \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{\left(1 - \frac{1}{2}\right)^2} - \frac{(1 + n \cdot 6) \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}}\\ & = 2 + 12(1 - \frac{1}{2^n}) - \frac{2(1 + 6n)}{2^n}\\ & = 2 + 12- \frac{12}{2^n} - \frac{(2 + 12n)}{2^n}\\ & = 14 - \frac{14 + 12n}{2^n}\\ & = \frac{14\times 2^n-14 + 12n}{2^n}\\ & = 2^{-n}(14\times 2^n-14 + 12n)\\ & = 2^{1-n}(7\times 2^n-7 + 6n)\end{align*}$$

This is the required sum.

< Question 23 & 24 Question 27 & 28 >