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Find the sum 12 + 4 + 4/3 + ...

Step by step, with the reason for every step · Series · Algebra 2

Step by step

The question

Question: find the sum 12 + 4 + 4/3 + and so on.

The dots at the end mean: it goes on forever.

12 + 4 + 43 + ...

1 Find the ratio

2 Look at the sums

1/3 is less than 1.

Each new term is smaller.

So the sums settle down near one number.

|r| = 13 < 1

Let's add the terms one at a time, and write each running sum: 12, 16, 17.33, and so on.

3 Use the sum rule

In the sum rule, r to the power n gets closer and closer to 0.

So it drops out, and the sum is the first term over 1 minus r.

S = a11 − r

Careful!

The terms go on forever, but the sum does not: each term is smaller than the one before.

Careful: forever, but the sum is not endless

Put in a₁ = 12 and r = 1/3.

S = 121 − 13

1 − 1/3 is 2/3.

So we divide by 2/3.

S = 12 ÷ 23

Dividing by 2/3 is times its flip, 3/2.

S = 12 × 32

So the sum is 18.

S = 18

4 Check

Let's check: we keep adding terms.

The running sums are 17.975, 17.992, 17.997.

sums: 17.778, 17.926, 17.975, 17.992, 17.997, ...

They get closer and closer to 18.

It works!

closer and closer to 18

Let's look back.

First: find the ratio.

Then: look at the sums.

Then: use the sum rule.

Last: check.

Answer
S = 18

Your turn

Try this one yourself, then check:

find the sum 12 + 6 + 3 + 32 + ...
Show the answer
24

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