Name this sequence

Calculus Level 2

$W=2+5x+9x^2+14x^3+20x^4+\cdots$

Given $|x|<1$ , find the value of the series $W$ .

Clarification : $2,5,9,14,20,\ldots$ follows a $2^{\text{nd}}$ -degree polynomial function, i.e. their second finite difference is a constant.

\frac{2-x}{(1+x)^3}

\frac{2+x}{(1+x)^3}

\frac{2+x}{(1-x)^3}

\frac{2-x}{(1-x)^3}

3 solutions

Maggie Miller
Aug 2, 2015

Edit: fixed this solution

$\displaystyle\frac{1}{x}\sum_{n=1}^{\infty}\left(\frac{(n+1)(n+2)}{2}-1\right)x^n=\frac{1}{x}\left( \frac{1}{2}\frac{d^2}{dx^2}\sum_{n=3}^{\infty}x^n\right)-\sum_{n=0}^{\infty}x^n$

$\displaystyle=\frac{1}{x}\left(\frac{1}{2}\frac{d^2}{dx^2}\left(\frac{x^3}{1-x}\right)\right)-\frac{1}{1-x}=\frac{1}{x}\left(\frac{1}{(1-x)^3}-1\right)-\frac{1}{1-x}$

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

Nope, there is no missing "x" in the answer. The correct answer is $\frac{2-x}{(1-x)^3}$ . Here is the brief solution:

$S=2+5x+9x^2+14x^3+20x^4+...$

Multiplying $S$ by x, we get

$Sx=2x+5x^2+9x^3+14x^4+20x^5+...$

Now subtracting $Sx$ from $S$ , we get

$S(1-x)=2+3x+4x^2+5x^3+6x^4+...$ (which is an AGP) -------(1)

Multiplying (1) by $x$ , and subtracting from $S(1-x)$ , we get

$S(1-x)^2=2+x+x^2+x^3+x^4+....=2+\frac{x}{1-x}=\frac{2-x}{1-x}$

$\implies S=\frac{2-x}{(1-x)^3}$ .

Sandeep Bhardwaj - 5 years, 10 months ago

Oh ok - I see now that I accidentally misindexed my series by 1, gaining an extra factor of $x$ .

Maggie Miller - 5 years, 10 months ago

So you solved the problem accidentally, right? ahahaa don't mind, i'm just kidding.

Please edit your solution accordingly and gain 1 upvote by me...

Sandeep Bhardwaj - 5 years, 10 months ago

@Sandeep Bhardwaj – Edited! :)

Maggie Miller - 5 years, 10 months ago

@Maggie Miller – Upvoted! :)

Sandeep Bhardwaj - 5 years, 10 months ago

Quiet ingenious enough, but I think you should mention the fact when you increment $n=3$ , from $n=1$ simultaneously we have to put $(n+2-2)(n+1-2)$ ......amateurs like me take time to grasp that.....

Arnav Das - 4 years, 9 months ago

Abu Bagwan
Mar 24, 2017

W=2+5x+9x^2+....------1 Wx=2x+5x^2+.....------2 Subtracting eq2 from eq 1 W(1-x)=2+x(3+4x+5x^2+......) W(1-x)=2+x{3/(1-x)+x/(1-x)^2} W=2-x/(1-x)^3

Syed Baqir
Sep 19, 2015

I love to Guess stuff::

We can guess here as well:

First formula for AGP for infinity is " $\rightarrow \frac {a}{1-r} + \frac {rd}{ (1-r)^{2} }$

Therefore Denominator must be negative therefore only three and four option could be correct

If we think a little the correct answer will seem to be third one if we solve a little : We may also think that x < 1

thank you!! this was so helpful

Sushma Vishwakarma - 3 years, 7 months ago

Name this sequence

3 solutions

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