My approach : I tried using integral calculus and using infinite geometric series..however it didn't match..any trick?
Asked
Active
Viewed 987 times
3
-
4hint: differentiate $f(x)=\frac 1{1-x}=\sum x^n$, at least for $|x|<1$. – lulu Oct 28 '16 at 15:47
-
1why are such clear duplicates not shut down immideatly? – tired Oct 28 '16 at 16:50
2 Answers
6
Hint: $$\frac{1}{1-x}=1+x+x^2+x^3+\dots$$ Differentiating with respect to $x$ and assuming $|x|<1$, which guarantees uniform convergence:
$$\dfrac{d}{dx}\left(\frac{1}{1-x}\right)=1+2x+3x^2+\dots.$$ (you should also check the radius of convergence of the resulting expression.)
An Alternative approach is based on the assumption of absolute convergence of the series:
$$S = 1 + x + x^2 + \dots$$ $$ + x + x^2 + \dots$$ $$ + x^2 + \dots$$ $$ \dots $$
MrYouMath
- 15,833
1
F = 1 + x + x^2 + ...
Your sum is F + x*F + x^2*F + x^3 * F ...
kotomord
- 1,814
-
Even I got this answer ...I expect a more appropriate answer...Anyways, thanks for ur time – Tanishka Marrott Oct 28 '16 at 16:26