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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?

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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.

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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 $\sum_{n=1}^\infty n x^n= {x \over (1-x)^2}$.

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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.

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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.

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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: $$\sum_{k=1}^{n} \frac{4k-2}{5^k}=4\sum_{k=1}^{n} \frac{k}{5^k}-2\sum_{k=1}^{n} \frac{1}{5^k}$$

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