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1+2x+3x^2+4x^3+.............nx^(n-1) Whats the sum of the

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1+2x+3x^2+4x^3+.............nx^(n-1) Whats the sum of the [#permalink] New post 20 Mar 2005, 05:29
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1+2x+3x^2+4x^3+.............nx^(n-1)

Whats the sum of the series?
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Re: PS: Series [#permalink] New post 20 Mar 2005, 05:55
gmat2me2 wrote:
1+2x+3x^2+4x^3+.............nx^(n-1)

Whats the sum of the series?


Let S = 1+2x+3x^2+4x^3+.............nx^(n-1) ... (1)

Multiply both sides by 'x'

xS = x + 2x^2+ 3x^3+.............nx^(n) ... (2)

Subtract (2) from (1)

S (1- x) = 1 + x + x^2 + ..... + x^(n-1) + nx^(n)

Except the last term, the rest is Geometric Progregssion with first term '1' and ratio 'x' with total 'n' terms

S(1-x) = 1 (1-x)/(1-x^n) + n X^n

S = 1/(1 - x^n) + nx^n/(1-x)

I hope I haven't done any mistake

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 [#permalink] New post 20 Mar 2005, 12:38
So the final formula would be the answer? Also, where did you find this problem?
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 [#permalink] New post 20 Mar 2005, 13:29
I agree with ketanm.
This can also be solved by differentiatingthe series w.r.t x but I guess it'd be a longer process
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Re: 1+2x+3x^2+4x^3+.............nx^(n-1) Whats the sum of the [#permalink] New post 21 Jan 2013, 13:32
In advance, I'm sorry for refreshing such an old post but I'm not quite sure If the solution ketanm had written is right.
ketanm wrote:
S(1-x) = 1 (1-x)/(1-x^n) + n X^n

The sum of geometric series is 1(1-x^n)/(1-x).
ketanm wrote:
S = 1/(1 - x^n) + nx^n/(1-x)

We can't divide by 1-x because x may equal 1.

What do u think about that?

Sry for my English, I am not a native speaker.
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Re: 1+2x+3x^2+4x^3+.............nx^(n-1) Whats the sum of the [#permalink] New post 21 Jan 2013, 20:11
Expert's post
gmat2me2 wrote:
1+2x+3x^2+4x^3+.............nx^(n-1)

Whats the sum of the series?


First up, it is not a GMAT relevant question at all. Still, if you want to solve it for intellectual purposes,

\(S = 1+2x+3x^2+4x^3+.............nx^{n-1}\)
\(xS = x+2x^2+3x^3+4x^4+.............nx^n\)

\(S - xS = 1 + x + x^2 + x^3 + .... + x^{n-1} - nx^n\)

We know that \(1 + x + x^2 + .... + x^{n-1} = (1 - x^n)/(1-x)\)(GP with n terms and x cannot be 1)

\(S(1-x) = (1 - x^n)/(1-x) - nx^n\)
\(S = (1 - x^n)/(1-x)^2 - nx^n/(1-x)\)
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Re: 1+2x+3x^2+4x^3+.............nx^(n-1) Whats the sum of the   [#permalink] 21 Jan 2013, 20:11
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