Question

pls answer my question I have written the formula

Answer 1

\[\sum_{r=1}^{\inf}(2r-1)(9/11)^{r}\]

Answer 2

@hartnn I have converted 9/11^r to 9/2
But how to convert (2r-1)

Answer 3

@wio pls help

Answer 4

@hartnn pls help

Answer 5

@ParthKohli pls help

Answer 6

Isn't this an arithmetic-geometric series?

Answer 7

yes

Answer 8

Then apply the formula...?

Answer 9

i have converted 9/11^r to 9/2 then what to do

Answer 10

\[\sum_{1}^{\infty}(2r-1)(\frac{ 9 }{ 11 })^r=\\2 \sum_{1}^{\infty}(r)(\frac{ 9 }{ 11 })^r-\sum_{1}^{\infty}(-1)(\frac{ 9 }{ 11 })^r=\]

Answer 11

^ can you do it now? :)

Answer 12

no I didn't got what @amoodarya is trying to explain

Answer 13

@amoodarya can u writ in equation

Answer 14

pls

Answer 15

second is easy geometric
for the first look this
\[x=\frac{ 9 }{ 11 }\\ \sum_{1}^{\infty}r x^r =s=1x+2x^2+3x^3+4x^4+....\\a=1+x+x^2+x^3+x^4+...=\frac{ 1 }{ 1-x }\\a'=1+2x+3x^2+4x^3+...=(\frac{ 1 }{ 1-x })'=\frac{ 1 }{ (1-x)^2 }\\x a' =x+2x^2+3x^3+... =\frac{ x }{ (1-x)^2 }\]

Answer 16

now @ the end put x=9/11

Answer 17

ok

Answer 18

\[2 \sum_{1}^{\infty}rx^r + \sum_{1}^{\infty}x^r=\\2 (\frac{ x }{ (1-x)^2 })+(x+x^2+x^3+...)\\2 (\frac{ x }{ (1-x)^2 })+(\frac{ x }{ 1-x })\\x=\frac{ 9 }{ 11 }\]

Answer 19

@paki ammodarya has written \[\sum_{r=1}^{\infty}(2r-1)(9/11)^r=\]

Answer 20

got it thnx for ur support @paki @ParthKohli @amoodarya