Question

Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it diverges. Summation from n equals 0 to infinity (StartFraction 9 Over 7 Superscript n EndFraction plus StartFraction 3 Over 5 Superscript n EndFraction )

Answer 1

The first four terms of the series are

`(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})`

`\sum_(n=0)^\infty \frac9{7^n}+(3)/(5^n)` = 14.25

Explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

`\sum_(k=0)^\infty a(r)^k`

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is
`\sum_(k=0)^\infty a(r)^k=(a)/(1-r)`

Given series,

`\sum_(n=0)^\infty \frac9{7^n}+(3)/(5^n)`

`=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......`

The first four terms of the series are

`(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})`

Let

`S_n=\sum_(n=0)^\infty (9)/(7^n)` and
`t_n=\sum_(n=0)^\infty (3)/(5^n)`

Now for
`S_n`,

`S_n=9+\frac97+(9)/(7^2)+\frac9{7^3}+.......`

`=\sum_(n=0)^\infty9(\frac 17)^n`

It is a geometric series.

The common ratio of
`S_n` is
`\frac17`

The sum of the series

`S_n=\sum_(n=0)^\infty (9)/(7^n)`

`=(9)/(1-\frac17)`

`=(9)/(\frac67)`

`=(9* 7)/(6)`

=10.5

Now for
`t_n`

`t_n= 3+\frac35+(3)/(5^2)+\frac3{5^3}+.......`

`=\sum_(n=0)^\infty3(\frac 15)^n`

It is a geometric series.

The common ratio of
`t_n` is
`\frac15`

The sum of the series

`t_n=\sum_(n=0)^\infty (3)/(5^n)`

`=(3)/(1-\frac15)`

`=(3)/(\frac45)`

`=(3* 5)/(4)`

=3.75

The sum of the series is
`\sum_(n=0)^\infty \frac9{7^n}+(3)/(5^n)`

=
`S_n+t_n`

=10.5+3.75

=14.25