# Find the sum to $n$ terms of the following series

Find the sum to $n$ terms of the following series: $$\dfrac {2}{5}+\dfrac {6}{5^2}+\dfrac {10}{5^3}+\dfrac {14}{5^4}+………$$

My Attempt: Let $$S_n=\dfrac {2}{5} + \dfrac {6}{5^2}+\dfrac {10}{5^3}+\dfrac {14}{5^4}+……+\dfrac {4n-6}{5^{n-1}}+\dfrac {4n-2}{5^n}$$ Also, $$\dfrac {1}{5} S_n=\dfrac {2}{5^2}+\dfrac {6}{5^3}+\dfrac {10}{5^4}+\dfrac {14}{5^5}+……+\dfrac {4n-6}{5^n}+\dfrac {4n-2}{5^{n+1}}$$

How do I solve further?

Isn't there any general method to solve such problems?

This is known as Arithmetrico-geometric sequence.

Suppose $$S_n =\sum_{k=1}^n [a+(k-1)d]r^{k-1}$$

then

$$S_n = \frac{a-(a+(n-1)d)r^n}{1-r}+\frac{dr(1-r^{n-1})}{(1-r)^2}$$

Here $a=2, d=4, r=\frac15$

\begin{align}S_n &= \frac{2-(4n-2)0.2^n}{0.8}+\frac{0.8(1-0.2^{n-1})}{0.8^2} \\ &= \frac{2-(4n-2)0.2^n}{0.8}+\frac{(1-0.2^{n-1})}{0.8} \\ &=\frac{3-(4n-2)0.2^n-5\cdot 0.2^n}{0.8}\\ &=\frac{3-(4n-3)0.2^n}{0.8}\\ &=\frac{15-(4n-3)0.2^{n-1}}{4}\end{align}

The derivation of the formula is exactly what you did, just subtract the two expressions.

The general term is ${4n-2 \over 5^n}$ so you can separate, compute $\sum_n {n\over 5^n}$ and $\sum_n {1\over 5^n}$ separately and then add the appropriate multiples to get the answer.

Note that since $\sum_{n=0}^\infty x^n= {1 \over 1-x}$, for $|x|<1$, we can differentiate to get $\sum_{n=1}^\infty n x^{n-1}= {1 \over (1-x)^2}$, or $\sum_{n=1}^\infty n x^n= {x \over (1-x)^2}$.

You're on the right track. Subtracting the two expressions you have, $$\frac45S_n = \frac4{5^2} + \frac4{5^3} + \dots + \frac4{5^n} - \frac{4n-2}{5^{n+1}}$$

Putting aside the last term, that expression should look more familiar to you...

In general, if something looks almost like a geometric series (or some other series you know how to sum), you want to manipulate it so that it is the thing you know.

You can subtract the two equations you got in such a way that you get a geometric progression.

$S_n - \dfrac {1}{5} S_n= \dfrac {2}{5} + \dfrac {6}{5^2}+\dfrac {10}{5^3}+\dfrac {14}{5^4}+……+\dfrac {4n-6}{5^{n-1}}+\dfrac {4n-2}{5^n} -( \dfrac {2}{5^2}+\dfrac {6}{5^3}+\dfrac {10}{5^4}+\dfrac {14}{5^5}+……+\dfrac {4n-6}{5^n}+\dfrac {4n-2}{5^{n+1}})$

$\dfrac{4}{5} S_n = \dfrac {2}{5} + 4 \sum_{i=2}^n \dfrac{1}{5^i} - \dfrac{4n-2} {5^(n+1)}$

Then you can sum the geometrical progression using regular formulae and you have an expression in n.

Hint:

You can treat your series as a combination of power and geometric series. See the following examples for $r\ne 0$ for power series and geometric series, respectively: $$\sum_{k=0}^{n} (kr^k)=\frac{r-(n+1)r^{n+1}+nr^{n+2}}{(r-1)^2}$$ $$\sum_{k=0}^{n} r^k=\frac{1-r^{n+1}}{1-r}$$ Note that your series can be written as: $$\sum_{k=1}^{n} \frac{4k-2}{5^k}=4\sum_{k=1}^{n} \frac{k}{5^k}-2\sum_{k=1}^{n} \frac{1}{5^k}$$