Summation: How to Solve

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
= 5a₁ + 5a₂ + 5a₃ + ... + 5a_n
= 5(a₁ + a₂ + a₃ + ... + 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.

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.

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 = n² + 1

So n ∑ a_k = n(n² + 1).

n(n² + 1) = n³ + n

So n³ + n is the answer.