Question

Sum of the following series

Answer 1

\[1+\frac{ 2 }{ 2 }+\frac{ 3 }{ 2^{2} }+\frac{ 4 }{ 2^{3} }.....\to n terms\]

Answer 2

\[\large{\sum_{n=1}^{n}\cfrac{n}{2^{n-1}}}\]

Answer 3

You are basically asking for this sequence ^^^

Answer 4

yes

Answer 5

It is much easier to find the sum once you are able to find out a general term

Answer 6

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
\[\large{\sum_{i=1}^{n}\cfrac{i}{2^{i-1}}}\]
\(\color{blue}{\text{End of Quote}}\)

Little correction :)

Answer 7

after this , how to proceed

Answer 8

this is AGP series

Answer 9

Yeah

Answer 10

Well I remember one method for such sequences lets see if it works :)

Answer 11

\[\large{S_n = 1 +\cfrac{2}{2} + \cfrac{3}{2^2} + ...}\tag{1}\]

\[\large{\cfrac{1}{2}S_n = 0 + \cfrac{1}{2} + \cfrac{2}{2^2} + \cfrac{3}{2^3} + ...}\tag{2}\]

Answer 12

Yeah this will work :D

Equation 1 - equation 2

Answer 13

We get:

\[\large{S_n(1-\cfrac{1}{2}) = 1 + \cfrac{1}{2} + \cfrac{1}{2^2} + ...+\cfrac{1}{2^{n-1}}-\cfrac{n}{2^n}}\]

\[\large{\cfrac{S_n}{2} +\cfrac{n}{2^n} = 1 + \cfrac{1}{2} + \cfrac{1}{2^2} + ...+\cfrac{1}{2^{n-1}}}\]

Answer 14

yesss

Answer 15

Now you can solve it easily. RHS is a GP.

Answer 16

I got it i think so , thanks

Answer 17

Your welcome :)

Answer 18

So would it be
\[\frac{ 0.5(0.5^{n}-1 )}{ -0.5 }\]

Answer 19

@vishweshshrimali5