Question

How do you find the sum of `Sigma (3^n+4^n)/5^n` from n is `[0,oo)`?

Answer 1

`sum_(n=0)^oo (3^n+4^n)/5^n = 15/2`

Explanation:

`sum_(n=0)^oo (3^n+4^n)/5^n=sum_(n=0)^oo (3^n/5^n + 4^n/5^n)`
`" "= sum_(n=0)^oo 3^n/5^n + sum_(n=0)^oo 4^n/5^n`
`" "= sum_(n=0)^oo (3/5)^n + sum_(n=0)^oo (4/5)^n`

The 1st sum is a GP with 1st term `a=1`, and common ratio `r=3/5`
The 2nd sum is a GP with 1st term `a=1`, and common ratio `r=4/5`

Using the GP formula: `S_n=a/(1-r)`, we get:

`sum_(n=0)^oo (3^n+4^n)/5^n = (1/(1-3/5)) + (1/(1-4/5))`
`" "= (1/(2/5)) + (1/(1/5))`
`" "= 5/2 + 5`
`" "= 15/2`