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Q 75.

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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.

(5k=120k+3)

The sum of this sequence is 1110.

See the step by step solution

Step by Step Solution

Step 1. Write the given information. 

The sum of sequence:

(5k+3)k=120

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,

(5k+3)k=120 =(5k)k=120+(3)k=120

&

(5k)k=120=5(k)k=120

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,

5k k=120=5×20(20+1)25k k=1n=5×20(21)2 5k k=1n=5×2105k k=1n=1050

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,3 k=120=3×203 k=120=60

Step 5. Now, add the two sums from Step 3 and Step 4.

From Step 3,

5k k=1n=1050

From Step 4,

3 k=120=60

So,

localid="1646750229919" (5k+3)k=120 =(5k)k=120+(3)k=120(5k+3)k=120=1050+60(5k+3)k=120=1110

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