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4 years ago

Sourabh Singh
IIT Patna
2104 Points


4 years ago
Mohammed Saleem Mohiuddin
23 Points
							We can as well use the method of infinite sum of an arithmeticogemetric series.Here A.P. is 1,2,3,4.... and G.P.is 1,1/2,1/2^2,1/2^3........a=1,d= 1, r=1/2,  Expression to  find infinite sum= a/(1-r)  +dr/(1-r)^2

4 years ago
mycroft holmes
272 Points
							We can see that the general term of the expression is $\frac{n}{2^{n-1}} = \frac{n+1}{2^{n-2} } - \frac{n+2}{2^{n-1} }$ i.e. it can be written in the form f(n) – f(n+1) where $f(n) = \frac{n+1}{2^{n-2} }$ Hence we have $\sum_{n=1}^{\infty} \frac{n}{2^{n-1}} =\sum_{n=1}^{\infty} \left(f(n)-f(n+1) \right ) = 4 - \lim_{n \rightarrow \infty}\frac{n+2}{2^{n-1}}$ $=\boxed{4}$

4 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions