Exercise 26 Page 711

We are given a finite arithmetic series in summation notation and want to find its sum. ∑^(10)_(k=0) (6k-1) To calculate the sum, we need to find the first and last terms. Let's substitute 1 and 10 for k in 6k-1.

a_k=6k-1
a_1=6( 1)-1	a_(10)=6( 10)-1
a_1=5	a_(10)=59

Now that we know a_1=5 and a_(10)=59, we will calculate the sum of the series substituting a_1=5, a_k=59, and k=10 in the formula for the sum of a finite arithmetic series.

S_k=k/2(a_1+a_k)

SubstituteValues

Substitute values

S_(10)=10/2( 5+ 59)

▼

Evaluate right-hand side

CalcQuot

Calculate quotient

S_(10)=5(5+59)

AddTerms

Add terms

S_(10)=5(64)

Multiply

Multiply

S_(10)=320

The sum of the terms a_1 to a_(10) of the series is 320. However, if we pay close attention to the given summation notation, we can see that our series starts at k= 0. This means that we want to calculate the sum from k=0 to k=10. ∑^(10)_(k= 0) (6k-1) Since we already found the sum of ten of the terms in the series, we will now calculate the first term, a_0, and add its value to the sum.

a_k=6k-1

Substitute

k=

a_()deepskyblue0=6( )-1

▼

Evaluate

Multiply

Multiply

a_0=0-1

SubTerm

Subtract term

a_0=- 1

Finally, we can find the sum of the terms indicated in the given summation notation. ∑^(10)_(k=0) (6k-1) = 320+(- 1)=319