We are given a finite series in summation notation and want to find its sum.

k = 3 \sum 6 (5 k - 2)

Since the formula is a linear function of

k,

we know that the series is arithmetic. To calculate the sum we need to find the

first

and

last

terms. Let's substitute

1

and

6

for

k

in

5 k - 2 .

$a_{k} = 5 k - 2$
$a_{1} = 5 (1) - 2$	$a_{6} = 5 (6) - 2$
$a_{1} = 3$	$a_{6} = 28$

Now that we know

a_{1} = 3

and

a_{6} = 28,

we will calculate the sum of the series by substituting

a_{1} = 3,

a_{n} = 28,

and

n = 6

in the formula for the sum of a finite arithmetic series.

S_{n} = \frac{n}{2} (a_{1} + a_{n})

SubstituteValues

Substitute values

S_{6} = \frac{6}{2} (3 + 28)

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

S_{6} = 3 (3 + 28)

AddTerms

Add terms

S_{6} = 3 (31)

Multiply

S_{6} = 93

The sum of the first six terms of the series is

93 .

However, if we pay close attention to the given summation notation we can see that our series starts at

n = 3 .

This means that we only want to calculate the sum from

k = 3

to

k = 6 .

k = 3 \sum 6 (5 k - 2)

Since we already found the sum of all the terms, we will now calculate the sum of the first two terms and find the difference. We will start by finding the first term.

a_{1} = 5 k - 2

Substitute

$k = 1$

a_{1} = 5 (1) - 2

▼

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

a_{1} = 5 - 2

SubTerm

Subtract term

a_{1} = 3

Let's find the second term.

a_{2} = 5 k - 2

Substitute

$k = 2$

a_{1} = 5 (2) - 2

▼

Evaluate right-hand side

Multiply

a_{1} = 10 - 2

SubTerm

Subtract term

a_{1} = 8

Having calculated the first two terms, we can find their sum.

S_{2} = a_{1} + a_{2}

SubstituteII

$a_{1} = 3$ , $a_{2} = 8$

S_{2} = 3 + 8

AddTerms

Add terms

S_{2} = 11

Finally, we can find the sum of the terms indicated in the given summation notation.

k = 3 \sum 6 (5 k - 2) = 93 - 11 = 82

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