Find the sum 1 + 2 + 4 + ... + 64 Step by Step

Step by step

The question

Question: find the sum 1 + 2 + 4 + and so on + 64.

1 + 2 + 4 + ... + 64

1 Find the ratio

2 Count the terms

The rule from the sequences lesson.

The last term is term n.

a_n = 1 × 2^{n − 1}

So 1 times 2 to the power (n − 1) is 64.

1 × 2^{n − 1} = 64

To remove times 1, we divide by 1.

64 divided by 1 is 64.

2^{n − 1} = 64

Which power of 2 is 64?

Multiply 2 by itself 6 times: we get 64.

So the power is 6.

2 × 2 × 2 × 2 × 2 × 2 = 64

So n − 1 = 6.

The power counts the jumps.

n − 1 = 6

Add 1 for the first term: n = 7 terms.

n = 7

Careful!

The power 6 counts the jumps.

There is one more term than jumps: 7 terms.

Careful: 6 jumps, so 7 terms (not 6)

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 = 1, r = 2 and n = 7.

S₇ = 1(2⁷ − 1)2 − 1

2 to the power 7 is 128.

S₇ = 1(128 − 1)1

128 − 1 is 127.

And over 1 is just itself.

S₇ = 1 × 127

So the sum is 127.

S₇ = 127

4 Check

Let's check: we add the 7 terms one by one.

Add: 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127

We get 127 again.

It works!

Let's look back.

First: find the ratio.

Then: count the terms.

Then: use the sum rule.

Last: check.

Answer