Question

Find the sum of the geometric series:
8 + 4 + 2 + 1 + ...

Answer 1

down to zero?

Answer 2

umm it doesn't say it just says....

Answer 3

\[\sum_{0}^{3}2^{n}= (1 - 2^{4})/(1- 2)\]

Answer 4

\[r=\frac{ 4 }{ 8}=\frac{ 1 }{ 2 }<1,a=8\]
\[S \infty=\frac{ a }{ 1-r }\]
substitute the value of a and r
get the value of S infty

Answer 5

a = 8
r = 1/2

S = a/(1-r)

Note: this formula only works if |r| < 1, which is true in this case (since |1/2| < 1 is true)

S=16

Answer 6

ok thanks guys:)

Answer 7

My Pleasure.

Answer 8

how abt this one?

Answer 9

17/.7

Answer 10

why would it be over .7?

Answer 11

\[\sum_{i=1}^{\infty}0.3^{i} = 1/(1-.3)\]

Answer 12

That should have been
\sum_{i=0}^{\infty}0.3^{i} = 1/(1-.3)

Answer 13

\[sum_{i=1}^{\infty}0.3^{i} = 1/(1-.3)\]

Answer 14

then what abt the 17???

Answer 15

That's just a factor multiplied times the whole sum, nothing special.

Answer 16

so multiply 1/.7 times 17

Answer 17

like this, 17(.3^0 + .3^1 + .3^2 + ...)

Answer 18

so 170/7?

Answer 19

yes, same thing, that's just 17/.7 * 1 = 17/.7 * 10/10

Answer 20

for those that don't like mixing decimals and fractions

Answer 21

ok thnks :)

Answer 22

not a problem