Simple Sequence

We are asked to find the sum of the fractions, $x =1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...$

Multiply both sides by 2 we obtain, $2x =2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...$

Notice RHS is $2 + x$ , thus , $2x = 2 + x$ $\boxed{x = 2}$

It is basically asking for the sum of an infinite geometric sequence. The formula is $\displaystyle \sum_{i=1}^\infty = \frac{a_{i}}{1-r}=\frac{1}{1- \frac{1}{2}}=\frac{1}{\frac{1}{2}}=\boxed{2}$

Instead of multiplying by 2, I divided it by 2 ..

X = 1+ 1/2 + 1/4 .... Div by 2 X/2 = 1/2 + 1/4....

X=1-X/2 X/2=1 X=2

S= a/(1-r) here a= initial term that mean 1 and r=common ratio among the terms that mean 1/2. Now if we put the value of a and r we can get Summation of this infinite series is, S=2. Answer is 2

$Given\quad That\\ 1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } ...................\\ Let\quad h=1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } ...................\\ Multi[lying\quad both\quad sides\quad by\quad "2"\\ 2h=2+1+\frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +\frac { 1 }{ 16 } .................\\ =>\quad 2h=2+h\quad or\\ =>\quad \boxed { h=2 }$

its an infinite GP with a=1 d=1/2 Sum of infinite GP is a/(1-r)=1/(1-1/2)=2

according to the given question we have a = 1 ( first term ) and b = 1/2 ( second term ) and we know that r = second term /first term = 1/2 /1 =1/2 put the value of " a " and " r " in the formula of sum of an infinite geometric sequence S = a /1 - r = 1 / 1 - 1/2 = 1 / 2 -1 /2 = 2 /1 = 2