A ride to infinity.....

Algebra Level 2

What is the sum of:

1 2 + 1 8 + 1 32 + 1 128 + . . . \large \color{#20A900} \frac {1}{2} + \frac {1}{8} + \frac {1}{32} + \frac {1}{128} + ...

if you added an infinite number of terms.

1 3 \frac {1}{3} 3 4 \frac {3}{4} 2 3 \frac {2}{3} 1 2 \frac {1}{2}

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2 solutions

Syed Hamza Khalid
Oct 15, 2017

If we notice clearly, this is a geometric sequence with the common ratio as 1 4 = r \frac{1}{4} = r . Therefore since r r is between 1 -1 and 1 1 , the sum to infinity exists. Using the formula of sum to infinity:

a 1 r where ’a’ is the first term \large \color{#3D99F6} \frac{a}{1 - r} \text{where 'a' is the first term}

So we can substitute the values in our equation:

= 0.5 1 0.25 = 0.5 0.75 = 2 3 \large \color{magenta}= \frac{0.5}{1 - 0.25} \\ \\ \large \color{magenta}= \frac{0.5}{0.75} \\ \\ \large \color{#EC7300} \boxed {= \frac{2}{3}}

Chew-Seong Cheong
Oct 17, 2017

S = 1 2 + 1 8 + 1 32 + 1 128 + 1 4 S = 1 8 + 1 32 + 1 128 + 1 512 + 3 4 S = 1 2 S = 2 3 \begin{aligned} S &= \frac 12 + \frac 18 + \frac 1{32} + \frac 1{128} + \cdots \\ \frac 14 S &= \frac 18 + \frac 1{32} + \frac 1{128} + \frac 1{512} + \cdots \\ \frac 34 S &= \frac 12 \\ \implies S &= \boxed{\dfrac 23} \end{aligned}

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