Question

Please help me find the sum for the following geometric series. (click to see)

Answer 1

\[\sum_{k=1}^{\infty} \frac{ 4^{k+2} }{ 7^{k-1}}\]

Answer 2

spose we have a geometric series
\[G=1+r+r^2+r^3+...+r^n\]
a simple solution to this is to subtract rG from G
\[~~~~~~G=1+r+r^2+r^3+...+r^n\\~~~~-rG=-r-r^2-r^3-...-r^n-r^{n+1}\]----------------------------------------\[(1-r)G=1+0+0+0+....+0-r^{n+1}\]
therefore\[G=\frac{1-r^{n+1}}{1-r}\]
ideally this setup has the top ending at n-1, so thenumerator is 1-r^n

Answer 3

lets pull out the constant, so that we only have the ^k parts left

Answer 4

\[\sum_{k=1}^{\infty} \frac{ 4^{k+2} }{ 7^{k-1}}\]
\[\sum_{k=1}^{\infty} \frac{ 4^{k} ~4^2 }{ 7^{k}~ 7^{-1}}\]
\[\sum_{k=1}^{\infty} (\frac{4}{7})^k~16*7\]

Answer 5

notice that the 112 pulls out and we are left with

G = r+r^2+r^3+r^4+....+r^k
-rG = -r^2- .................-r^k-r^(k+1)
--------------------------------------
(1-r)G = r - r^(k-1)

G = \(\Large \frac{r - r^{(k-1)}}{1-r}\)
the limit as k to inf
G = \(\Large \frac{r - r^{inf}}{1-r}\)will depend on if 4/7 < 1

Answer 6

if so, then r^inf = 0, if |r| < 1

Answer 7

ok thank you for the help
i can find the sum from here

Answer 8

:) have fun