badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass

                        

how to add to infinity 1+2/2+3/2^2+4/2^3......Answer given is 4

4 years ago

Answers : (3)

Sourabh Singh
IIT Patna
askIITians Faculty
2104 Points
							243-1352_capture.PNG
						
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
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

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


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details