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Q 76.

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Precalculus Enhanced with Graphing Utilities
Found in: Page 809
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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Short Answer

In Problems 71-82, find the sum of each sequence.

(3k=126k-7)

The sum of this sequence is 871.

See the step by step solution

Step by Step Solution

Step 1. Write the given information. 

The sum of sequence:

(3k-7)k=126

Step 2. Use the properties of sequence.

Using the properties of sequence:

(ak-bk)k=1n =(ak)k=1n-(bk)k=1n&(cak)k=1n=cakk=1n

So,

(3k-7)k=126 =(3k)k=126-(7)k=126

&

(3k)k=126=3(k)k=126

Step 3. Use the formula for sum of sequences of n real numbers.

Using formula for summation is:

k k=1n=1+2+...+nk k=1n=n(n+1)2

So,

role="math" localid="1646750162806" 3k k=126=3×26(26+1)23k k=126=3×26(27)2 3k k=126=3×3513k k=126=1053

Step 4. Use the formula for sum of sequence. 

Using formula for summation is:c k=1n=c+c+...+cc k=1n=cn , c-real numberSo,7 k=126=7×267 k=126=182

Step 5. Now, subtract Step 3  from Step 4.

From Step 3,

3k k=126=1053

From Step 4,

7 k=126=182

So,

(3k-7)k=126 =(3k)k=126-(3)k=126(3k-7)k=126=1053-182(3k-7)k=126=871

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