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$\begingroup$

Here, $a=1, d=2, b=1, r=x$ \begin{align} S_n&= \frac{ab}{1-r}+\frac{bdr(1-r{^n}^{-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^n}{1-r}\\ S_n&=\frac 1 {1-x}+\frac{ 2x(1-x{^n}^{-1})} {(1-x)^2}-\frac{[1+(n-1)(2)]x^n} {1- x}\\ &= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x.x{^n}^{-1}}{(1-x)^2}-\frac{[1+2n-2]x^n}{(1-x)}\\ &= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x^n}{(1-x)^2}-\frac{[2n-1]x^n}{(1-x)}\\ \end{align}

Is it correct. I have not got the answer, please show me how to move to this answer without skipping any line $\frac {1-3x} {(1-x)^2}+\frac {2x^n}{(1-x)^2}-\frac{(2n-1)x^n}{(1-x)}$

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    In the title, you ask for "the $n$-th of the series" and, in the body of the text, you calculate the sum of $n$ terms of the series. – 2017-01-21
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    What is the question? – 2017-01-21
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    https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence – 2017-01-21
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    @Bernard Massé Sorry for the mistake, question edited. – 2017-01-21

3 Answers 3

6

Another way

\begin{align*} 1+3x+5x^2+7x^3+\cdots+(2n-1)x^{n-1}&=(1+2x+3x^2+4x^3+\cdots+nx^{n-1})\\&+x(1+2x+3x^2+\cdots+(n-1)x^{n-2})\\ &=\frac{d}{dx}(x+x^2+x^3+\cdots+x^n)\\&+x\frac{d}{dx}(x+x^2+x^3+\cdots+x^{n-1})\\ &=\frac{d}{dx}\left(\frac{x^{n+1}-x}{x-1}\right)+x\frac{d}{dx}\left(\frac{x^{n}-x}{x-1}\right)\\ \end{align*}

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    Can you tell is my process correct or not. – 2017-01-21
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    What is $S_n$ or $a,b,d?$ – 2017-01-21
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    $n$ should stand for the total number of terms. I guess the asker intend to take the limit that $n$ approaches infinity. $a$ denotes the initial term of the series, $d$ denotes the common difference between the coefficients of the consecutive powers of $x$, and $r$ denotes the common ratio between consecutive terms (ignoring the coefficients). I don't know about $b$ though. – 2017-01-21
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    @Behrouz Maleki, Here a= 1st term of AP and d= common difference, b= first term of GP and r= common ratio of GP. Hope you understand now. – 2017-01-22
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    @changer I see .Your solution is correct. – 2017-01-22
4

Here's a pretty simple approach:

$ S = 1 + 3x + 5x^2 + 7x^3 + \cdots + (2t + 1)x^t \\ \implies xS = x + 3x^2 + 5x^3 + \cdots + (2t - 1)x^t + (2t + 1)x^{t+1} \\ $

where $t = n - 1$. Subtracting $xS$ from $S$,

$$ (1 - x)S = 1 + 2x + 2x^2 + 2x^3 + \cdots + 2x^t + (2t+1)x^{t+1}. $$

Replacing $t$ by $n - 1$,

$ (1-x)S = [-1 + (2n - 1)x^n] + 2(1 + x + x^2 + \cdots + x^{n-1})\\ \implies (1-x)S = [-1 + (2n - 1)x^n] + 2\frac{1 - x^n}{1 - x} \\ \implies S = \frac{(2n - 1)x^n}{1 - x} - \frac{1}{1-x} + 2\frac{1-x^n}{(1-x)^2}. $

By the way, it's not important for the condition $0

0

Here's how I did it:

let $S = 1 + 3x + 5x^2 + 7x^3 + ....$

By expansion, $S = \displaystyle\sum_{n=0}^{\infty}(2n+1)x^n .... (1)$

but similarly, $S = \displaystyle\sum_{n=1}^{\infty}(2n-1)x^{n-1} ....(2)$

Examining $(2)$, $(2n-1)x^{n-1} = \displaystyle\frac{1}{x}(2n-1)x^n$ $ = \displaystyle\frac{1}{x}(2n+1-2)x^n = \displaystyle\frac{1}{x}[(2n+1)x^n - 2x^n]$

So $(2)$ becomes

$S = \displaystyle\frac{1}{x}\sum_{n=1}^{\infty}[(2n+1)x^n - 2x^n]$

But using $(1)$, we get

$S = \displaystyle\frac{1}{x}(S-1)- \frac{2}{x}\sum_{n=1}^{\infty} x^n ....(3)$

(making term limits of $(1)$ as 1 and infinity and subtracting the first term.)

Now $\displaystyle\sum_{n=1}^{r} x^n = x + x^2 + x^3 + .... = x(1 + x + x^2 + ...)$ is the geometric series $\displaystyle\frac{x(1-x^r)}{1-x}$.

Hence $(3)$ becomes

$S = \displaystyle\frac{1}{x}(S-1)- \displaystyle\frac{2(1-x^r)}{1-x}$

which makes $S = \displaystyle\frac{x+1-2x^{r+1}}{(1-x)^2}$ for $x>1$

or $S = \displaystyle\frac{x+1}{(1-x)^2}$ if $0