how do i find the sum of a series from n=1 to infinity of ((3^n)+5)/ (4^n) ?

Question

safari321 · Answer

you will 
$$((3^{1}+5) \div(4^{1})$$
$$((3+5)\div4$$
$$8\div4$$
$$=2$$

anonymous · Answer

i got that far but i need help finding r

anonymous · Answer

seperate to 2 pieces 3^n/4^n  + 5/4^n

anonymous · Answer

right. i also separated them but i don't know how to find r after that

anonymous · Answer

first part has a = 3/4 and r = 3/4   second part has a = 5/4 and r = 1/4

mertsj · Answer

However (3^2+5)/4^2=14/16=7/8

mertsj · Answer

The sum is 4 2/3

anonymous · Answer

okay so mathwiz707, what do i do to create the sum with those two separate parts?

anonymous · Answer

first part sum = 3/4/(1 -3/4) and second has sum 5/4/(1-1/4)  ... now just add the 2 parts

anonymous · Answer

oh okay. so is that 42/3 that you wrote as the answer?

anonymous · Answer

3 + 15 = 18 ...

anonymous · Answer

wait, what does ^ that have to do with this problem?

anonymous · Answer

means exponent

anonymous · Answer

when i added the two sums i got 14/3 because you add 3+ (5/3) not 3+15

anonymous · Answer

thanks for the help! appreciate it