Just A Sum

Algebra Level 1

$1+ \dfrac{3}{4} + \dfrac{9}{16} + \dfrac{27}{64} + \cdots = \, ?$

The answer is 4.

4 solutions

Hung Woei Neoh
Jun 28, 2016

Relevant wiki: Geometric Progression Sum

Notice that $1+\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots$ is a sum of a geometric progression to infinity, where $a=1$ and $r=\dfrac{3}{4}$

Therefore, we can use the formula $S_{\infty} =\dfrac{a}{1-r}$ to calculate this

$1+\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots\\ =\dfrac{1}{1 - \frac{3}{4}}\\ =\dfrac{1}{\frac{1}{4}}\\ =\boxed{4}$

What do you think @Gabe Smith 's image is doing there / how does it relate to your solution?

Eli Ross Staff - 4 years, 11 months ago

Let's see...

Let $S = 1+\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots$

The first two triangles are similar triangles, therefore the ratio of their sides are the same

$\dfrac{1+\frac{3}{4}+\frac{9}{16}+\frac{27}{64}+\ldots}{1} = \dfrac{\frac{3}{4}+\frac{9}{16}+\frac{27}{64}+\ldots}{\frac{3}{4}}\\ \dfrac{3}{4}\left(1+\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots\right) = \dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots\\ \dfrac{3}{4}S = -1+1+ \dfrac{3}{4}+\dfrac{9}{16}+\dfrac{27}{64}+\ldots\\ \dfrac{3}{4}S = -1+S\\ S-\dfrac{3}{4}S = 1\\ S=\dfrac{1}{1-\frac{3}{4}}=\boxed{4}$

We will get the same solution by equating the ratios of the sides of the two triangles

Hung Woei Neoh - 4 years, 11 months ago

Kexin Zheng
Jun 28, 2016

This problem can be seen as a geometric series with $a_1=1$ and common ratio $r=\frac{3}{4}$ Thus, the sum is $\frac{a_1}{1-r}=\frac{1}{1-\frac{3}{4}}$ $=\boxed{4}$

Geoff Pilling
Dec 1, 2018

The diagram helps a lot. Clearly the side length of the bottom length is the sum.

And the slope of the hypotenuse is $\dfrac{1}{4}$ .

So, a glance at the triangle leads us to conclude that the series adds up to $\boxed4$ .

A Former Brilliant Member
Mar 23, 2017

Just A Sum

The answer is 4.

4 solutions

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