Question

$\large\textsf{Find the sum to n terms and also } S_{\infty}\\\\\dfrac{3}{5} + \dfrac{5}{15} +\dfrac{7}{45}+ \dfrac{9}{135}+ ............$

Answer 1

hope it is helpful to you

Answer 2

This one is called arithmetico geometric series and is given by :

a , (a + d) r , (a + d)r² ..... and so on .

Well we have to make it like this and then we need to find Sum of the terms . Well we have :

$S_n=\dfrac{a}{1-r}+\dfrac{dr[1-r^{n-1}]}{(1-r)^2}-\dfrac{nth\:term\:of\:numerator\times r^n}{1-r}$

BIG NOTE :

This is for r < 1 which is in this case .

Well we are not done yet because it is not an arithmetico geometric series yet  :\

Let the sum be S till n terms.

$S=\dfrac{3}{5}+\dfrac{5}{15}+....\\\\\implies 5S=3+\dfrac{5}{3}+.....\\\\\implies 5S=\dfrac{a}{1-r}+\dfrac{dr[1-r^{n-1}]}{(1-r)^2}-\dfrac{nth\:term\:of\:numerator\times r^n}{1-r}\\\\\implies 5S=\dfrac{a}{1-r}+\dfrac{dr[1-r^{n-1}]}{(1-r)^2}-\dfrac{(a+[n-1]d)r^n}{1-r}\\\\\implies 5S=\dfrac{3}{1-1/3}+\dfrac{2\times 1/3[1-(1/3)^{n-1}]}{(1-1/3)^2}-\dfrac{(3+[n-1]2)(1/3)^n}{1-1/3}\\\\\implies 5S=\dfrac{3\times 3}{2}+\dfrac{3}{2}[1-(\dfrac{1}{3})^{n-1}]-\dfrac{(2n+1)(1/3)^{n})}{2/3}$

$5S=\dfrac{3\times 3}{2}+\dfrac{3}{2}[1-(\dfrac{1}{3})^{n-1}]-\dfrac{(2n+1)(1/3)^{n})}{2/3}\\\\\implies 5S=\dfrac{9}{2}+\dfrac{3}{2}-\dfrac{3}{2}\times 3^{1-n}-\dfrac{3(2n+1)}{2.3^n}\\\\\implies 5S=\dfrac{9+3}{2}-\dfrac{3}{2}\times 3^{1-n}-(\dfrac{2n+1}{2})\times 3^{1-n}\\\\\implies 5S=\dfrac{12}{2}-\dfrac{3}{2}\times \dfrac{3}{3^n}-\dfrac{3}{3^n}\dfrac{(2n+1)}{2}\\\\\implies 5S=6-\dfrac{3}{3^n}(\dfrac{3}{2}+\dfrac{2n+1}{2})\\\\\implies 5S=6-\dfrac{3}{3^n}(\dfrac{2n+4}{2}$

$\implies 5S=6-\dfrac{3}{3^n}(n+2)\\\\\implies 5S=3[2-\dfrac{n+2}{3^n}]\\\\\implies \boxed{S=\dfrac{3}{5}[2-\dfrac{n+2}{3^n}]}$

Now the formula for infinite terms can be derived by limits. The answer will be too big so I am not deriving .

$S=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}\\\\\implies S=\dfrac{3}{5}+\dfrac{5}{15}+......\infty\\\\\implies 5S=3+\dfrac{5}{15}+....\infty\\\\\implies 5S=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}\\\\\implies 5S=\dfrac{3}{1-\dfrac{1}{3}}+\dfrac{\dfrac{2}{3}}{(1-\dfrac{1}{3})^2}\\\\\implies 5S=\dfrac{9}{2}+\dfrac{\dfrac{2}{3}}{\dfrac{4}{9}}\\\\\implies 5S=\dfrac{9}{2}+\dfrac{3}{2}\\\\\implies 5S=\dfrac{12}{2}\implies 6\\\\\implies S=\boxed{\dfrac{6}{5}}$