# Summation: How to Solve

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

## Example 1

### Example

### Solution

See the first term 5a_{k}.

5 is a coefficient.

So 5 gets out from the sigma.

The variable of the sigma is k.

So a_{k} is in the sigma.

This is true because

∑ 5a_{k}

= 5a_{1} + 5a_{2} + 5a_{3} + ... + 5a_{n}

= 5(a_{1} + a_{2} + a_{3} + ... + a_{n})

= 5 ∑ a_{k}.

Sigma Notation

Next, see the next term 2b_{k}.

+2 is a coefficient.

So +2 gets out from the sigma.

The variable of the sigma is k.

So b_{k} is in the sigma.

So ∑ (5a_{k} + 2b_{k})

= 5 ∑ a_{k} + 2 ∑ b_{k}.

∑ a_{k} = 7

∑ b_{k} = -3

So 5 ∑ a_{k} + 2 ∑ b_{k}

= 5⋅7 + 2⋅(-3).

5⋅7 = 35

+2⋅(-3) = -6

35 - 6 = 29

So 29 is the answer.

## Example 2

### Example

### Solution

See the first term 2a_{k}.

2 is a coefficient.

So 2 gets out from the sigma.

The variable of the sigma is k.

So a_{k} is in the sigma.

Next, see -7.

-7 is a constant.

Then it becomes -7n.

This is true because

∑ (-7) means

adding -7 n times: -7n.

So ∑ (2a_{k} -7)

= 2 ∑ a_{k} - 7n.

∑ a_{k} = 5

So 2 ∑ a_{k} - 7n = 2⋅5 - 7n.

2⋅5 = 10

Arrange the terms in descending order.

So -7n + 10 is the answer.

## Example 3

### Example

### Solution

See the term na_{k}.

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 a_{k} is in the sigma.

∑ a_{k} = n^{2} + 1

So n ∑ a_{k} = n(n^{2} + 1).

n(n^{2} + 1) = n^{3} + n

So n^{3} + n is the answer.