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find the taylor series of 5/(1-5x)^2
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Consider the antiderivative, \[\int\frac{5}{(1-5x)^2}\,dx=\frac{1}{1-5x}+C\] Recall that for \(|x|<1\), you have \[\frac{1}{1-x}=\sum_{k=0}^\infty x^k\] So, if \(f(x)=\dfrac{5}{(1-5x)^2}\), and \(F(x)=\displaystyle\int f(x)\,dx\), then \[F(x)=\sum_{k=0}^\infty (5x)^k+C\] Differentiating will give you the series for \(f(x)\).
thanks, do u have to change the starting index k to 1 also?
ok i think so :)
It wouldn't matter. \[\sum_{k=0}^\infty k x^{k-1}=0+\underbrace{1+2x+3x^2+\cdots}_{\text{starting at }k=1}=\sum_{k=1}^\infty kx^{k-1}\]
kk i see :D
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