Write 3 + 6 + 12 + 24 using sigma notation Step by Step

Step by step

Question: write 3 + 6 + 12 + 24 using sigma notation.

3 + 6 + 12 + 24

The rule from the sequences lesson, with i for the term number: the first term, times 2 to the power (i − 1).

r = 2: a_i = 3 × 2^{i − 1}

So the rule is 3 × 2 to the power (i − 1).

a_i = 3 × 2^{i − 1}

i counts the terms: 1, 2, 3, 4.

There are 4 terms.

So i starts at 1 and stops at 4.

i = 1, 2, 3, 4

Careful!

The stop number on Σ is the last i, 4.

It is not the last term, 24.

Careful: not Σ_{i = 1}²⁴3 × 2^{i − 1} ✗

So we write: the sum for i = 1 to 4 of 3 × 2 to the power (i − 1).

= Σ_{i = 1}⁴3 × 2^{i − 1}

Let's check: put i = 1, 2, up to 4 into 3 × 2 to the power (i − 1).

We get each term of the list.

i = 1 to 4: 3 + 6 + 12 + 24

That is the sum we started with.

It works!

Let's look back.

First: find the rule.

Then: count the terms.

Then: write it with Σ.

Last: check.

Answer

= Σ_{i = 1}⁴3 × 2^{i − 1}