Sum from i = 1 to 6 of (12 - 2i) Step by Step

Step by step

The question

Question: work out the sum for i = 1 to 6 of (12 − 2i).

Σ_{i = 1}⁶(12 − 2i)

1 Read the sigma

Σ is the Greek letter sigma.

It means: add up.

Start with i = 1, and stop at i = 6.

That is 6 − 1 + 1: 6 terms.

For each i we work out 12 − 2i.

Σ = add up

start: i = 1, stop: i = 6

6 − 1 + 1 = 6 terms

Careful!

The + 12 is inside the bracket.

So it is added in every term, 6 times, not just once.

Careful: not Σ_{i = 1}⁶−2i + 12 ✗

2 Write the terms

i = 1: we put 1 in place of i.

We get 10.

i = 1: 12 − 2(1) = 12 − 2 = 10

i = 2 gives 8.

i = 2: 12 − 2(2) = 12 − 4 = 8

i = 3: 12 − 2(3) = 12 − 6 = 6

i = 4: 12 − 2(4) = 12 − 8 = 4

i = 5: 12 − 2(5) = 12 − 10 = 2

And i = 6 gives 0.

That is the last one.

i = 6: 12 − 2(6) = 12 − 12 = 0

3 Add them

Now we add the 6 terms.

Σ_{i = 1}⁶(12 − 2i) = 10 + 8 + 6 + 4 + 2 + 0

So the sum is 30.

= 30

4 Check

Let's check with the pairs from the sequences lesson.

The terms go up by the same amount, so first plus last, 10 + 0, times 6 terms, over 2.

Pairs: 6 × (10 + 0)2 = 6 × 102 = 30

We get 30 again.

It works!

Let's look back.

First: read the sigma.

Then: write the terms.

Then: add them.

Last: check.

Answer