# Summation: How to Solve

How to solve the given summation by using the summation properties: 3 examples and their solutions.

## Example 1

### Solution

See the first term 5ak.

5 is a coefficient.
So 5 gets out from the sigma.

The variable of the sigma is k.
So ak is in the sigma.

This is true because
∑ 5ak
= 5a1 + 5a2 + 5a3 + ... + 5an
= 5(a1 + a2 + a3 + ... + an)
= 5 ∑ ak.

Sigma Notation

Next, see the next term 2bk.

+2 is a coefficient.
So +2 gets out from the sigma.

The variable of the sigma is k.
So bk is in the sigma.

So ∑ (5ak + 2bk)
= 5 ∑ ak + 2 ∑ bk.

∑ ak = 7
∑ bk = -3

So 5 ∑ ak + 2 ∑ bk
= 5⋅7 + 2⋅(-3).

5⋅7 = 35
+2⋅(-3) = -6

35 - 6 = 29

## Example 2

### Solution

See the first term 2ak.

2 is a coefficient.
So 2 gets out from the sigma.

The variable of the sigma is k.
So ak is in the sigma.

Next, see -7.

-7 is a constant.
Then it becomes -7n.

This is true because
∑ (-7) means

So ∑ (2ak -7)
= 2 ∑ ak - 7n.

∑ ak = 5

So 2 ∑ ak - 7n = 2⋅5 - 7n.

2⋅5 = 10

Arrange the terms in descending order.

So -7n + 10 is the answer.

## Example 3

### Solution

See the term nak.

The variable of the sigma is k, not n.
So think n as a constant, in this case.
Then n gets out from the sigma.
And ak is in the sigma.

∑ ak = n2 + 1

So n ∑ ak = n(n2 + 1).

n(n2 + 1) = n3 + n

So n3 + n is the answer.