0
$\begingroup$

Solution

Here $a=1, d=3, b=1, r=\frac15$

\begin{align} S_n &=\frac 1 {1-1/5}+\frac{ 3(\frac{1}{5})[1-(\frac{1}{5}){^n}^{-1}} {(1 1/5)^2}-\frac {[1+(n-1)(3)](\frac 15)^n} {1-1/5} \\ &= \frac 54+\frac {15} {16}[1-(1/5){^n}^{-1}]-\frac54[3n-2](1/5){^n}\\ &= \frac 54+\frac {15} {16}-\left[\frac {15} {16}+\frac{3n-2}{4}\right]\left(\frac 15\right){^n}^{-1} \end{align}

Just added 3rd line from book but i did not understand. How $$\left[\frac {15} {16}+\frac{3n-2}{4}\right]\left(\frac 15\right){^n}^{-1}$$ this came from the 2nd line.

  • 0
    The first term of the series is $1$ or $\frac{1}{5^2}$?. What is $S_1$? – 2017-01-15
  • 1
    https://en.m.wikipedia.org/wiki/Arithmetico-geometric_sequence – 2017-01-15
  • 0
    @JoseA132 This series in A.G.P (arithmetic o geometric progression). So first term of arithmetic progression is $a=1$ and common difference $d=3$. And the first term of geometric progression is $b=1$ and common ratio $r= \frac15$ – 2017-01-15

2 Answers 2

2

It is an A.G.P. i.e. combination of an A.P. and a G.P.

Its $n^{th}$ term would be$$T_n =\frac{1+3(n-1)}{5^{n-1}}$$ as you can see the numerator is $n^{th}$ of the A.P. and denominator is $n^{th}$ term of the G.P.

Let $$S = 1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+...+\frac{1+3(n-1)}{5^{n-1}}$$

Multiplying both sides by $\frac{1}{5}$ which is the common ratio

$$\frac S 5 = \frac15+\frac4{5^2}+\frac7{5^3}+...+\frac{1+3(n-2)}{5^{n-1}}+\frac{1+3(n-1)}{5^{n}}$$

From above equations $$S-\frac S5 = 1+\frac 35+\frac 3{5^2}+\frac 3{5^3}+...+\frac{3}{5^{n-1}}-\frac{1+3(n-1)}{5^{n}}$$ We observe a G.P. with first term $\frac 35$, common ratio $\frac15$ and no. of terms is $n-1$ leaving aside the first and the last term. $$\implies4\frac S5 = 1-\frac{1+3(n-1)}{5^{n}}+\frac{\frac 35((\frac 15)^{n-1}-1)}{\frac 15-1}$$ Simplify it further and that will be your answer.

0

The sum can be written as:

$$\sum_{k=0}^n \frac{1 + 3k}{5^k} = \sum_{k=0}^n \left( \frac15 \right)^k + 3 \sum_{k=0}^n \frac{k}{5^k} = \frac{1 - \left( \frac15\right)^{n+1} }{1-\frac15} + 3s_n(1/5)$$

where $s_n(x) = \sum_{k=0}^n kx^k$. We have: $$s_n(x) = x\frac{d}{dx} \sum_{k=0}^n x^k = x \frac{d}{dx} \frac{1-x^{n+1}}{1-x}= \frac{x(nx^{n+1} -(n+1)x^n +1)}{(1-x)^2}$$

Now compute.