Question

How do you find the sum of the finite geometric sequence of `Sigma 2(4/3)^n` from n=0 to 15?

Answer 1

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

`2[(4/3)^o + (4/3)^1 + (4/3)^2...(4/3)^15]`

Sum of any geometric finite geometric series is,

`= (a(r^n-1))/(r-1)` if `r!=1`

where,
`a =` first term of the series
`r =` common ratio of the series

substituting the values,

`2[(1((4/3)^16-1))/(4/3-1)]`

`2[((4^16-3^16)/3^16)/(1/3)]`

`2[(4^15-3^15)/3^15]`

Symplify and get the answer.

Answer 2

`sum_(n=0)^15 2(4/3)^n approx 592.647311`

Explanation:

Sum `=sum_(n=0)^15 2(4/3)^n = 2sum_(n=0)^15 (4/3)^n`

Now, `sum_(n=0)^15 (4/3)^n` is the sum of geometric progression, with 1st term `a=1` and common ratio `r=4/3`

The sum of a finite GP `= (a(1-r^n))/(1-r)` where `n` is the number of terms.

In our example, `a=1, r=4/3, n=16`

`:. sum_(n=0)^15 (4/3)^n = (1*(1-(4/3)^16))/(1-4/3)`

`approx -98.77455184xx -3`

`approx 296.3236555`

Sum `= 2 * sum_(n=0)^15 (4/3)^n`

`approx 2xx 296.3236555`

`approx 592.647311`