Respuesta :
Answer:
The first four terms of the series are
[tex](9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})[/tex]
[tex]\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}[/tex] = 14.25
Step-by-step explanation:
We know that
Sum of convergent series is also a convergent series.
We know that,
[tex]\sum_{k=0}^\infty a(r)^k[/tex]
If the common ratio of a sequence |r| <1 then it is a convergent series.
The sum of the series is [tex]\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}[/tex]
Given series,
[tex]\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}[/tex]
[tex]=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......[/tex]
The first four terms of the series are
[tex](9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})[/tex]
Let
[tex]S_n=\sum_{n=0}^\infty \frac{9}{7^n}[/tex] and [tex]t_n=\sum_{n=0}^\infty \frac{3}{5^n}[/tex]
Now for [tex]S_n[/tex],
[tex]S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......[/tex]
[tex]=\sum_{n=0}^\infty9(\frac 17)^n[/tex]
It is a geometric series.
The common ratio of [tex]S_n[/tex] is [tex]\frac17[/tex]
The sum of the series
[tex]S_n=\sum_{n=0}^\infty \frac{9}{7^n}[/tex]
[tex]=\frac{9}{1-\frac17}[/tex]
[tex]=\frac{9}{\frac67}[/tex]
[tex]=\frac{9\times 7}{6}[/tex]
=10.5
Now for [tex]t_n[/tex]
[tex]t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......[/tex]
[tex]=\sum_{n=0}^\infty3(\frac 15)^n[/tex]
It is a geometric series.
The common ratio of [tex]t_n[/tex] is [tex]\frac15[/tex]
The sum of the series
[tex]t_n=\sum_{n=0}^\infty \frac{3}{5^n}[/tex]
[tex]=\frac{3}{1-\frac15}[/tex]
[tex]=\frac{3}{\frac45}[/tex]
[tex]=\frac{3\times 5}{4}[/tex]
=3.75
The sum of the series is [tex]\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}[/tex]
= [tex]S_n+t_n[/tex]
=10.5+3.75
=14.25
