Question

If y=x+x2+x3....... Where |x|<1 prove that x=y/(1+y)

Answer 1

Heya!!!

Sum of infinite terms of G.P is given by

a/ ( 1 - r )

Where a is ist term and r is common ratio

_______________________________

=>

x + x² + x³ + x⁴ +.... Are in G.P becoz common ratio is same.

y = x/ ( 1 - x )

=>

y ( 1 - x ) = x

=>

y - yx = x

=>

y = x + yx

=>

x = y/( 1 + y )

Answer 2

Using the formula of sum of infinite GP we can easily prove $x=\dfrac{y}{y+1}$.

Step-by-step explanation:

The given equation is

$y=x+x^2+x^3+...$

Where |x|<1 .

We need to prove that $x=y(1+y)$.

y represents the sum of an infinite GP $x,x^2,x^3,...$. First term is x and common ratio is x.

The sum of an infinite GP is

$S_n=\dfrac{a}{1-r}$

where, a is first term and r is common ratio, |r|<1.

$y=\dfrac{x}{1-x}$

$y(1-x)=x$

$y-xy=x$

$y=x+xy$

$y=x(1+y)$

Divide both sides by (1+y).

$\dfrac{y}{1+y}=x$

$x=\dfrac{y}{y+1}$

Hence proved.

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