(2+4x+6x^2+8x^3+\cdots+(2n)x^{n-1})
[ S=2\sum_{k=1}^{n} k x^{k-1} ] Or (S=2+4x+6x^2+\cdots+2n x^{n-1})
Standard result: [ S=\dfrac{2\bigl[1-(n+1)x^n+nx^{n+1}\bigr]}{(1-x)^2} ]
Answer: (\dfrac{2[1-(n+1)x^n+nx^{n+1}]}{(1-x)^2})
11TH•Mathematics•Sequences And Series•Question 8
Overview
Sum to $n$ terms the series $2 + 4x + 6x^2 + 8x^3 + \cdots$ where $x \neq 1$.
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