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adell [148]
2 years ago
11

Stormer Company reports the following amounts on its statement of cash flow: Net cash provided by operating activities was $40,0

00; net cash used in investing activities was $14,800 and net cash used in financing activities was $17,400. If the beginning cash balance is $6,800, what is the ending cash balance?
Mathematics
1 answer:
Archy [21]2 years ago
8 0

Answer:

its D

Step-by-step explanation:

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Use sigma notation to represent the sum of the first seven terms of the following sequence -4,-6,-8.....
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Answer:Answer:

\sum\left {{7} \atop {1}} \right -n(3+n)

Step-by-step explanation:

Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;

The sum of an arithmetic series is expressed as S_n = \frac{n}{2}[2a+(n-1)d]

n is the number of terms

a is the first term of the sequence

d is the common difference

Given parameters

n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2

Required

Sum of the first seven terms of the sequence

S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 =  \frac{7}{2}[-8+(6)(-2)]\\\\S_7 =  \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70

The sum of the nth term of the sequence will be;

S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n =  \frac{-6n}{2} -  \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)

The sigma notation will be expressed as \sum\left {{7} \atop {1}} \right -n(3+n). <em>The limit ranges from 1 to 7 since we are to  find  the sum of the first seven terms of the series.</em>

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The <em>correct answer</em> is:


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