   Chapter 13.1, Problem 22E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 20-25, use the sum formulas I-V to express each of the following without the summation symbol. In Problems 20-23, find the numerical value. ∑ k = 1 50 ( 6 k 2 + 5 )

To determine

To calculate: The expression of k=150(6k2+5) without the summation symbol and find its value.

Explanation

Given Information:

The provided sum is k=150(6k2+5).

Formula used:

The sum of n numbers x1, x2, x3,,xn is given by

k=1nxk=x1+x2++xn

The value of the sum k=1n(xk+yk)=k=1nxk+k=1nyk.

The value of the sum k=1n1=n.

The value of the sum k=1ncxk=ck=1nxk.

The value of the sum k=1nk2=n(n+1)(2n+1)6.

Calculation:

Consider the sum k=150(6k2+5).

The sum of n numbers x1, x2, x3,,xn is given by k=1nxk=x1+x2++xn.

So, the k=150(6k2+5) can be expanded by

(6(1)2+5)+(6(2)2+5)+(6(3)2+5)++(6(50)2+5).

Recall that the value of the sum k=1n(xk+yk)=k=1nxk+k=1nyk.

Thus, the sum k=150(6k2+5) can be written as,

k=150(6k2+5)=k=1506k2+k=1505

Recall that the value of the sum k=1ncxk=ck=1nxk

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