How do I sum the series $\infty \sum n = 1 n {(\frac{3}{4})}^{n}$ $sum_(n=1)^oon(3/4)^n$ ?

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How do I sum the series $\infty \sum n = 1 n {(\frac{3}{4})}^{n}$ $sum_(n=1)^oon(3/4)^n$ ?

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Answer 1 · 2015-02-12T19:05:02

The answer is: 12.

First of all: "How much is the sum of a geometric series?".

$+ \infty \sum n = 0 r^{n} = \frac{a_{0}}{1 - r}$ $sum_(n=0)^(+oo)r^n=a_0/(1-r)$ , where r (such as $| r | < 1$ $|r|<1$ ) is called "ratio" ( $r = \frac{a_{k}}{a_{k - 1}}$ $r=a_k/a_(k-1)$ ) and $a_{0}$ $a_0$ is the first term of the series (in this case $a_{0} = r^{0} = 1$ $a_0=r^0=1$ ).

Let's list all the terms of the series:

$1*3/4 + 2*(3/4)^2+3*(3/4)^3+4*(3/4)^4+...$ , but we can also say:

$3/4+9/16+9/16+27/64+27/64+27/64+81/256+81/256+81/256+81/256+...$ ,

or, better:

$3/4+9/16+27/64+81/256+...+$

$9/16+27/64+81/256+...+$

$27/64+81/256+...+$

$81/256+...+$

$...$

The first sum is:

$sum_(n=1)^(+oo)(3/4)^n$ ,

The second sum is:

$sum_(n=2)^(+oo)(3/4)^n$ ,

The third sum is:

$sum_(n=3)^(+oo)(3/4)^n$ ,

The fourth sum is:

$sum_(n=4)^(+oo)(3/4)^n$ ,

$...$

So:

$sum_(n=1)^(+oo)n(3/4)^n=sum_(n=1)^(+oo)(3/4)^n+sum_(n=2)^(+oo)(3/4)^n+ sum_(n=3)^(+oo)(3/4)^n+sum_(n=4)^(+oo)(3/4)^n+...=$ .

$=3/4*1/(1-3/4)+9/16*1/(1-3/4)+27/64*1/(1-3/4)+81/256*1/(1-3/4)+...=$

$=3/4*4+9/16*4+27/64*4+81/256*4+...=$

$=4(3/4+9/16+27/64+81/256+...)=$

$=4*sum_(n=1)^(+oo)(3/4)^n=4*3/4*1/(1-3/4)=4*3/4*4=12$ .

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How do I sum the series $\infty \sum n = 1 n {(\frac{3}{4})}^{n}$ $sum_(n=1)^oon(3/4)^n$ ?

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How do I sum the series $\infty \sum n = 1 n {(\frac{3}{4})}^{n}$ $sum_(n=1)^oon(3/4)^n$ ?