Question:\( 1 + 2x + 3x^2+4x^3+...+nx^{n-1}\): What is the sum of the series?

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

Answer: B

Approach Solution (1):

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

Multiply by x on both the sides, we have:

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 + \(x^2+x^3+...+x^{n-1}\) = \(\frac{(1-x^n)}{(1-x)}\) (GP with n terms and x cannot be 1)

S (1 – x) = \(\frac{(1-x^n)}{(1-x)}\) – \(nx^n\)

S = \(\frac{(1-x^n)}{(1-x)}-\frac{nx^n}{(1-x)}\)

Approach Solution (2):

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

Multiply both sides by ‘x’

xS = x + \(2x^2+3x^3+4x^4+...+nx^n\) --- (2)

Subtract (2) from (1), we have:

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

Except the last term, the rest Geometric Progression with first term ‘1’ and ratio ‘x’ with the total ‘n’ terms

S (1 – x) = \(\frac{(1-x^n)}{(1-x)}\) – \(nx^n\)

S = \(\frac{(1-x^n)}{(1-x)}-\frac{nx^n}{(1-x)}\)

“What \( 1 + 2x + 3x^2+4x^3+...+nx^{n-1}\): What is the sum of the series?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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