how to find the sum from 0 to infinity (2^(K+3))/3^K

Question

anonymous · Answer

$$\sum_{0}^{\infty} 2^{k+3}/3^{k}$$

amistre64 · Answer

2^k   2^3/3^k

amistre64 · Answer

got it

amistre64 · Answer

$$\sum2^3(\frac{2}{3})^k\to 8\sum (\frac{2}{3})^k $$

amistre64 · Answer

the sum part goes to 1/(1-2/3)

amistre64 · Answer

1/(1/3) = 3

8*3 = 24

anonymous · Answer

great explanation! Got it man!

amistre64 · Answer

yw

anonymous · Answer

what if the denominator goes from 3^k to 3^(K+3) and then numerator became 1

amistre64 · Answer

same concepts, just different numbers really

anonymous · Answer

$$\sum_{0}^{\infty} 1/3^{k+3}$$

amistre64 · Answer

$$B^{a+b}\to B^aB^b$$

that allow you to factor off the constant and use the rest as a ratio

amistre64 · Answer

1/3^3 sum (1/3)^k

amistre64 · Answer

the sum of a geometric sequence is:$$\sum \frac{1-r^n}{1-r}\to \frac{1-r^{inf}}{1-r}$$
when r is less than 1, r^inf goes to zero

amistre64 · Answer

well, when |r|<1 ...

anonymous · Answer

true
now I'm struggling on this question $$\sum_{0}^{\infty} 3^{k-1}/4^{3k+1}$$
i try to use your method, but didn't work
I have no idea how to make the denominator good

amistre64 · Answer

well, lets start by stripping off the boring things :)

amistre64 · Answer

$$ \sum_{0}^{\infty} 3^{k-1}/4^{3k+1}$$

$$ \sum_{0}^{\infty} 3^{k}3^{-1}/4^{3k}4^{1}$$

$$ 1/12\sum_{0}^{\infty} 3^{k}/4^{3k} $$

$$ 1/12\sum_{0}^{\infty} 3^{k}/(4^3)^k $$

$$ 1/12\sum_{0}^{\infty} (3/64)^k $$

amistre64 · Answer

now r=3/64

anonymous · Answer

oh ya!!