How to solve this series which is in both AP and GP?

Question

how to solve this series:

Comments

$E(X) = 2*(\frac{1}{2})^{1} + 3*(\frac{1}{2})^{2} + 4*(\frac{1}{2})^{3} + ......\infty$

It is the combination of AP and GP... Terms 2,3,4,... are in AP and Terms $\frac{1}{2} , (\frac{1}{2})^{2},...$ are in GP..

To solve these type of series , there is a fixed procedure...We multiply by the common ratio of GP (here it is 1/2) in  both LHS and RHS..while multiplying by common ratio in RHS , we have to write by leaving space of one term and then subtract 2nd equation from 1st..

We repeat this procedure until we get the GP in RHS...

So, Here ,

$E(X) \;\;\;\;\;= \;\;\;\;\;\;2*(\frac{1}{2})^{1} + 3*(\frac{1}{2})^{2} + 4*(\frac{1}{2})^{3} + 5*(\frac{1}{2})^{4}+......\infty$

$\frac{1}{2}*E(X) = \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2*(\frac{1}{2})^{2} + 3*(\frac{1}{2})^{3} + 4*(\frac{1}{2})^{4} + ......\infty$

Now, vertically Subtract $2^{nd}$ equation from $1^{st}$ equation.

$(1-\frac{1}{2})*E(X)= 2*(\frac{1}{2})^{1}+ 1*(\frac{1}{2})^{2} + 1*(\frac{1}{2})^{3} + 1*(\frac{1}{2})^{4} + ......\infty$

$(\frac{1}{2})*E(X)= 2*(\frac{1}{2})^{1}+$ GP with first term as $(\frac{1}{2})^{2}$ and common ratio as 1/2...

$(\frac{1}{2})*E(X)$= 1 + $ \frac{(\frac{1}{2})^{2}}{(1-\frac{1}{2})}$

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@ ankitgupta.1729  Thank you very much for the perfect explanation!

Answer 1

This is an AGP.

Given,

$$S = 2 \times \frac{1}{2} + 3 \times \frac{1}{4} \dots $$

Multiply the series by $1/2$,

$$S/2 = 2 \times \frac{1}{4} + 3 \times \frac{3}{8} \dots$$

Subtracting the second equation from the first,

$$S/2 = 2 \times \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \dots $$

This is now an infinite GP, whose sum is known.

The final answer will be:

$S/2 = 1 + \frac{1/4}{3/4}$

Comments

@ goxul I am not getting the subtraction part, can you please clarify it a bit more.

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When you subtract the second from the first, arrange it in such a way that the first term of the second series aligns with the second term of the first series. When you subtract them (for example, in the first instance, it'll be $3 \times 1/4 - 2 \times 1/4$, which will give you $1/4$. Notice that all the fractions become singular. I hope this is clear.

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thank you very much goxul.