Question

How do you compute `sum_"n=1"^oo(4^n+5^n)/6^n`?

Answer 1

`sum(n=1)^oo(4^n+5^n)/6^n=5+2=7`

Explanation:

We can split this up into

`sum_(n=1)^oo[4^n/6^n+5^n/6^n]=sum_(n=1)^oo[(4/6)^n+(5/6)^n]`

Recall that you can split summations up across sums and differences.

`sum_(n=1)^oo(4/6)^n+sum_(n=1)^oo(5/6)^n`

`(4/6)^n=(2/3)^n`.

`sum_(n=1)^oo(2/3)^n+sum_(n=1)^oo(5/6)^n`

Recall that if we have `sum_(n=0)ar^n, |r|<1,` then the value is given by `a/(1-r).`

Our series are pretty close to the above form, but we need to start at zero. We can do this by adding `1` to all `n`.

`sum_(n=0)^oo(2/3)^(n+1)+sum_(n=0)^oo(5/6)^(n+1)`

Recalling that `a^(b+c)=a^ba^c, (2/3)^(n+1)=2/3(2/3)^n, (5/6)^(n+1)=5/6(5/6)^n`

Thus, we have

`sum_(n=0)^oo(2/3)(2/3)^n+sum_(n=0)^oo(5/6)(5/6)^n`

`sum_(n=0)^oo2/3(2/3)^n=(2/3)/(1-(2/3))=(2/3)/(1/3)=2/3*3=2`

`sum_(n=0)^oo5/6(5/6)^n=(5/6)/(1-(5/6))=(5/6)/(1/6)=5/6*6=5`

Thus, we have

`sum_(n=1)^oo(4^n+5^n)/6^n=5+2=7`

Answer 2

`sum_(n=1)^oo \ (4^n+5^n)/6^n = 7`

Explanation:

We seek the sum:

`S = sum_(n=1)^oo \ (4^n+5^n)/6^n`

We can manipulate as follows:

`S = sum_(n=1)^oo \ (4^n/6^n + 5^n/6^n)`

`\ \ = sum_(n=1)^oo \ ( (4/6)^n + (5/6)^n)`

`\ \ = sum_(n=1)^oo \ (2/3)^n + sum_(n=1)^oo \ (5/6)^n`

The first sum is a GP with `a=2/3` and `r=2/3`
The second sum is a GP with `a=5/6` and `r=5/6`

Using the standard GP formula:

`s_oo = a/(1-r) \ \ \ \ |r| lt 1`

Then, we can write the summation as:

`S = (2/3)/(1-2/3) + (5/6)/(1-5/6)`

`\ \ = (2/3)/(1/3) + (5/6)/(1/6)`

`\ \ = 2+ 5`

`\ \ = 7`