Question

Find the first five terms of the sequence of partial sums. (Round your answers to four decimal places.) [infinity] (−5)n + 1 n!

Answer 1

Answer:

25.0000 + -37.5000 + 66.6667 + -63.5416 + 66.6667

Step-by-step explanation:

The actual formatting of the question has been attached to this response.

From the question,

Let the sequence of terms be $b_{n}$ i.e

$b_{n}$ = $\frac{(-5)^{n+1} }{n!}$

Let the sequence of partial sums be $S_{n}$ i.e

$S_{n}$ = s₁ + s₂ + s₃ + . . . + sₙ

Therefore the first five terms of the sequence of partial sums will be S₅ i.e

S₅ =  s₁ + s₂ + s₃ + s₄ + s₅

Where;

s₁ = b₁

s₂ = b₁ + b₂ = s₁ + b₂

s₃ = b₁ + b₂ + b₃ = s₂ + b₃

s₄ = b₁ + b₂ + b₃ + b₄ = s₃ + b₄

s₅ = b₁ + b₂ + b₃ + b₄ + b₅ = s₄ + b₅

Where;

b₁ can be found by substituting n = 1 into equation (i) as follows;

$b_{1}$ = $\frac{(-5)^{1+1} }{1!}$

$b_{1}$ = 25

$b_{1}$ = 25.0000

Recall that

s₁ = b₁

∴ s₁ = 25.0000 to 4 decimal places

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b₂ can be found by substituting n = 2 into equation (i) as follows;

$b_{2}$ = $\frac{(-5)^{2+1} }{2!}$

$b_{2}$ = -62.5

$b_{2}$ = -62.5000

Recall that

s₂ = s₁ + b₂

∴ s₂ = 25.000 + -62.5000 = -37.5000

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b₃ can be found by substituting n = 3 into equation (i) as follows;

$b_{3}$ = $\frac{(-5)^{3+1} }{3!}$

$b_{3}$ = 104.1667

Recall that

s₃ = s₂ + b₃

∴ s₃ = -37.5000 + 104.1667 = 66.6667

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b₄ can be found by substituting n = 4 into equation (i) as follows;

$b_{4}$ = $\frac{(-5)^{4+1} }{4!}$

$b_{4}$ = -130.2083

Recall that

s₄ =  s₃ + b₄

∴ s₄ = 66.6667 + -130.2083 = -63.5416

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b₅ can be found by substituting n = 5 into equation (i) as follows;

$b_{5}$ = $\frac{(-5)^{5+1} }{5!}$

$b_{5}$ = 130.2083

Recall that

s₅ = s₄ + b₅

∴ s₅ = -63.5416 + 130.2083 = 66.6667

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Therefore, the first five terms of the partial sum is:

25.0000 + -37.5000 + 66.6667 + -63.5416 + 66.6667