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 March 25th, 2011, 07:41 AM #1 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 find the sum find the sum of this sequence $2x+4x^2+6x^3+8x^4+.............2nx^n$
 March 25th, 2011, 08:23 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: find the sum $1+x+x^2+x^3+...+x^n+x^{n+1}\quad=\quad\frac{1-x^{n+2}}{1-x}$ Start by differentiating both sides.
March 25th, 2011, 10:14 AM   #3
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Re: find the sum

Quote:
 Originally Posted by aswoods $1+x+x^2+x^3+...+x^n+x^{n+1}\quad=\quad\frac{1-x^{n+2}}{1-x}$ Start by differentiating both sides.
i dont get u
what does diffirentiation has to do with my sequence?

 March 25th, 2011, 10:42 AM #4 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: find the sum ok i get it now thanks buddy
 March 25th, 2011, 12:24 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: find the sum Using aswoods' helpful suggestion: (1) $1+x+x^2+x^3+\cdots+x^n+x^{n+1}=\frac{1-x^{n+2}}{1-x}$ Differentiating both sides: (2) $1+2x+3x^2+\cdots+nx^{n-1}+(n+1)x^n=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2}$ Subtract (1) from (2): $x+2x^2+3x^3+\cdots+(n-1)x^{n-1}+nx^n=\frac{1-x^{n+2}}{x-1}+\frac{(n+1)x^{n+2}-(n+2)x^{n+1}+1}{(x-1)^2}+x^{n+1}$ Combine right-side with common denominator and simplify: $x+2x^2+3x^3+\cdots+(n-1)x^{n-1}+nx^n=\frac{x$$x^n\(n(x-1)-1$$+1\)}{(x-1)^2}$ Multiply through by 2: $2x+4x^2+6x^3+\cdots+2(n-1)x^{n-1}+2nx^n=\frac{2x$$x^n\(n(x-1)-1$$+1\)}{(x-1)^2}$
March 25th, 2011, 05:57 PM   #6
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Re: find the sum

Hello, islam!

This can be solved algebraically.

Quote:
 $\text{Find the sum of this sequence: }\:2x\,+\,4x^2\,+\,6x^3\,+\,8x^4\,+\,\cdots\,+\,2n x^n$

$\begin{array}{cccccccccccccc}\text{We have:}=&S=&2x=&+=&4x^2=&+=&6x^3=&+=&8x^4=&+=&\cdots=&+=&2nx^n=&\\ \\ \\ \text{Multiply by }x:=&xS=$

[color=beige]. . [/color]$\text{We have: }\,\,\,\,(1\,-\,x)S \;=\;2x\underbrace{(1\,+\,x\,+\,x^2\,+\,x^3\,+\,\c dots\,+\,x^{n-1})}_{\text{geometric series}} \,-\,2nx^{n+1}$

[color=beige]. . . . . . . . . . .[/color]$(1\,-\,x)S \;=\;2x\,\cdot\frac{1\,-\,x^n}{1\,-\,x}\;-\;2nx^{n-1}$

[color=beige]. . . . [/color]$\text{Therefore: }\;\;\;\;\;S \;=\;2x\,\cdot\,\frac{1\,-\,x^n}{(1\,-\,x)^2} \;-\;\frac{2nx^{n+1}}{1\,-\,x}$

 March 25th, 2011, 10:58 PM #7 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: find the sum Don't forget the case x=1...
March 25th, 2011, 11:52 PM   #8
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Re: find the sum

Quote:
 Originally Posted by aswoods Don't forget the case x=1...
n(n + 1)

 March 26th, 2011, 01:24 AM #9 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: find the sum thank u aswoods ,markfl, soroban,chaz i am really grateful for ur help u all r really creative i wish u all great breakthroughs in math, wealth and happiness.

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