Question

sum of infinite GP
$1 + 2x + 3x { }^{2} + 4x {}^{3} + ....... \infty$
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Answer 1

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Answer 2

Answer:

⇒ $\frac{1}{(1-x)^2}$

Step-by-step explanation:

Given :

To find the sum of :

$1+2x+3x^2+4x^3+...\infty$

Solution :

We know that,

If G.P series given by,

$S = a + ar+ar^2+ar^3+...\infty$

The sum upto infinity terms is given by,

$a + ar+ar^2+ar^3+...\infty = \frac{a}{1-r}$ ....Identity

here,

the given series can be written as,

$1+2x+3x^2+4x^3+...\infty$

⇒ $1+(x+x)+(x^2 + x^2+x^2)+(x^3+x^3+x^3+x^3)+...\infty$

⇒ $(1+x+x^2+x^3+...\infty) + (x+x^2+x^3+...\infty)+(x^2+x^3+x^4+...\infty)+(x^3+x^4+x^5+...\infty)+...$

⇒ $[(1+x+x^2+x^3+...\infty) + x(1+x+x^2+x^3+...\infty)+x^2(1+x+x^2+x^3+...\infty)+....]$

⇒ $(1+x+x^2+x^3+...\infty)(1+x+x^2+x^3+...\infty)$

and we can define the single series's summation by using the identity,

So,

$(1+x+x^2+x^3+...\infty) = \frac{1}{1-x}$

and hence,

⇒ $(1+x+x^2+x^3+...\infty)(1+x+x^2+x^3+...\infty)= (\frac{1}{1-x})(\frac{1}{1-x})$

⇒ $\frac{1}{(1-x)^2}$