Question

k=1 sigma k tends to infinity 4^(k+1)/7^(k-1) ?
this is my qustion how can calculte the sum of this series..?

Answer 1

\[\sum_{k=1}^\infty{4^{k+1}\over7^{k-1}}\]correct?

Answer 2

yup.:)

Answer 3

you need the actual sum, not just to know that it converges?

Answer 4

yup i find that converges or not..? but how.?

Answer 5

\[\sum_{k=1}^\infty{4^{k+1}\over7^{k-1}}=\sum_{k=1}^\infty16(\frac47)^{k-1}\]now it's a simple geometric series

Answer 6

the sum of any geometric series of the form\[\sum_{k=1}^\infty ar^{k-1}=\frac a{1-r}~~~~if~~~~~|x|<1\]

Answer 7

bro can u explain ..how..?

Answer 8

16 sigma ..? how..?

Answer 9

\[4^{k+1}=4^2\cdot4^{k-1}\]

Answer 10

\[=16(4^{k-1})\]

Answer 11

oho ok ok i understand bro

Answer 12

got it from here?

Answer 13

yes |r|<1 and a=16 me right.. na..?

Answer 14

yep

Answer 15

so the sum is...?

Answer 16

i calciule just 2 mints

Answer 17

16/21 ?

Answer 18

not what I got

Answer 19

what is r ?

Answer 20

r=4/7

Answer 21

so what is 1-r ?

Answer 22

3/7

Answer 23

so what is
16/(3/7) ?

Answer 24

112/3

Answer 25

there we go :)

Answer 26

thanks..:)