Restating the equation required,

$x = \frac{1}{6} + \frac{1}{6^{2}} + \frac{1}{6^{3}} + + \frac{1}{6^{4}} + ...$

Multiplying both sides by 6 we obtain,

$6x = 1 + \frac{1}{6} + \frac{1}{6^{2}} + \frac{1}{6^{3}} + \frac{1}{6^{4}} + ...$

Notice that the RHS is equivalent to $1+x$ , which can be written as

$6x = 1 + x$

Solving for $x$ , $\boxed{ x = \frac{1}{5} = 0.2}$

This is a simple GP question and the formula to find sum of infinite terms is =A/(1-R) where A is the first term and R is the ratio of first and second terms basically second term / first term.

$Let\\ \qquad S=\frac { 1 }{ 6 } +\frac { 1 }{ { 6 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 3 } } +\frac { 1 }{ { 6 }^{ 4 } } +.....................\infty \\ Multiplying\quad Both\quad sides\quad by\quad "6",\quad we\quad have\\ \qquad 6S=1+\frac { 1 }{ 6 } +\frac { 1 }{ { 6 }^{ 2 } } +\frac { 1 }{ { 6 }^{ 3 } } +\frac { 1 }{ { 6 }^{ 4 } } +.....................\infty \\ =>6S=1+S\\ =>6S-S=1\quad =>\quad 5S=1\\ or\quad S=\frac { 1 }{ 5 } \\ THus\quad the\quad answer\quad is\quad \boxed { S=0.2 }$

Let $\frac {1}{6} +\frac {1}{6^2} ... = x$ . Then factor out the $\frac {1}{6}$ from the second term onwards. Thus you end up with $\frac {1}{6} + \frac {1}{6}(\frac {1}{6} +\frac{1}{6^2}+...) = x$ . Remember that you set the expression in the brackets as x. Thus you substitute x in for the expression in the brackets. You will end up with $\frac {1}{6} +\frac {1}{6}x = x$ . Solve for x and you will get 0.2.

By using sum of Geometric series formula S= a(1- r^n/1-r) where a= 1/6 , common ratio r= 1/6 and n= 4 the answer is 0.2

thanks for question

S=a/1-r
a=1/6 & r=1/6
so
S=1/5=0.2

Equation for infinite GP , S=a/(1-r). So, S=(1/6)/(1-(1/6))=0.2

Geometric series with a1 = 1/6 and r = 1/6

Sum = a1/(1-r)

= (1/6)/(1-1/6) = (1/6)/ (5/6) = 1/5 = .2

S = a/1 - r

so a =1/6 , b = 1/36 and r = b/a =1/36 /1/6 = we get r = 1/6 so , put the value of a , r in above formula S = 1/6 /1- 1/6 = 1/5 = 0 .2

infinite sum = a / 1- r, where a is first term and r is common ratio.

here a = 1/6 = r

infinite sum = (1 /6 ) / (1 - (1 / 6) ) = 1/ 5 = 0.2

lets add 1and subtract 1 then, 1+1/6+1/6square+.....-1 now using infinite gp 1/(1-1/6) -1 6/5-1 1/5 0.2