# Thread: rewrite as geometric series

1. ## rewrite as geometric series

Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
Thank you.

2. Originally Posted by cowboys111
Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
Thank you.
$\sum_{n=0}^{\infty}\frac{4^{n+2}}{7^{n-1}}=\sum_{n=0}^{\infty}\frac{16}{\frac{1}{7}}\bigg (\frac{4}{7}\bigg)^{n}=112\sum_{n=0}^{\infty}\bigg (\frac{4}{7}\bigg)^{n}=112\frac{1}{1-\frac{4}{7}}=261.33$

3. Originally Posted by cowboys111
Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
Thank you.
I think you're right on your geometric series as 112(4^k/7^k) but if a = 64 (the first value of your geometric series), then your lower limit of summation must be 1. Putting in 1 for k yields a = 64 for your first value.

Not sure but I think that's what it is.