Find the sum 3 + 6 + 12 + ... + 384 Step by Step

Step by step

The question

Question: find the sum 3 + 6 + 12 + and so on + 384.

3 + 6 + 12 + ... + 384

1 Find the ratio

2 Count the terms

The rule from the sequences lesson.

The last term is term n.

a_n = 3 × 2^{n − 1}

So 3 times 2 to the power (n − 1) is 384.

3 × 2^{n − 1} = 384

To remove times 3, we divide by 3.

384 divided by 3 is 128.

2^{n − 1} = 128

Which power of 2 is 128?

Multiply 2 by itself 7 times: we get 128.

So the power is 7.

2 × 2 × 2 × 2 × 2 × 2 × ... = 128

So n − 1 = 7.

The power counts the jumps.

n − 1 = 7

Add 1 for the first term: n = 8 terms.

n = 8

Careful!

The power 7 counts the jumps.

There is one more term than jumps: 8 terms.

Careful: 7 jumps, so 8 terms (not 7)

3 Use the sum rule

The sum rule from the sequences lesson (the S and rS trick): the first term, times r to the power n minus 1, over r minus 1.

S_n = a₁(rⁿ − 1)r − 1

Put in a₁ = 3, r = 2 and n = 8.

S₈ = 3(2⁸ − 1)2 − 1

2 to the power 8 is 256.

S₈ = 3(256 − 1)1

256 − 1 is 255.

And over 1 is just itself.

S₈ = 3 × 255

So the sum is 765.

S₈ = 765

4 Check

Let's check: we add the 8 terms one by one.

Add: 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384

= 765

We get 765 again.

It works!

Let's look back.

First: find the ratio.

Then: count the terms.

Then: use the sum rule.

Last: check.

Answer