Find the sum 5 + 15 + 45 + ... + 1215 Step by Step

Step by step

The question

Question: find the sum 5 + 15 + 45 + and so on + 1215.

5 + 15 + 45 + ... + 1215

1 Find the ratio

2 Count the terms

The rule from the sequences lesson.

The last term is term n.

a_n = 5 × 3^{n − 1}

So 5 times 3 to the power (n − 1) is 1215.

5 × 3^{n − 1} = 1215

To remove times 5, we divide by 5.

1215 divided by 5 is 243.

3^{n − 1} = 243

Which power of 3 is 243?

Multiply 3 by itself 5 times: we get 243.

So the power is 5.

3 × 3 × 3 × 3 × 3 = 243

So n − 1 = 5.

The power counts the jumps.

n − 1 = 5

Add 1 for the first term: n = 6 terms.

n = 6

Careful!

The power 5 counts the jumps.

There is one more term than jumps: 6 terms.

Careful: 5 jumps, so 6 terms (not 5)

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 = 5, r = 3 and n = 6.

S₆ = 5(3⁶ − 1)3 − 1

3 to the power 6 is 729.

S₆ = 5(729 − 1)2

729 − 1 is 728.

S₆ = 5 × 7282

So the sum is 1820.

S₆ = 1820

4 Check

Let's check: we add the 6 terms one by one.

Add: 5 + 15 + 45 + 135 + 405 + 1215

= 1820

We get 1820 again.

It works!

Let's look back.

First: find the ratio.

Then: count the terms.

Then: use the sum rule.

Last: check.

Answer