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

Answer:

The first four terms of the series are

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

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step 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 $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

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

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

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$    and     $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

    $=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

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

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

    $=\frac{9}{1-\frac17}$

    $=\frac{9}{\frac67}$

    $=\frac{9\times 7}{6}$

    =10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

    $=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

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

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

    $=\frac{3}{1-\frac15}$

    $=\frac{3}{\frac45}$

    $=\frac{3\times 5}{4}$

    =3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

                                        = $S_n+t_n$

                                       =10.5+3.75

                                       =14.25