Question

Find 10 partial sums of the series. (round your answers to five decimal places.) ∞ 15 (−4)n n = 1

Answer 1

Find 10 partial sums of the series. (Round your answers to five decimal places.)

∞

15 /(−4)n

Do calculations based on answer above (i.e. 15/(-4)^1 + 15/(-4)^2+...

1

-3.75000

Correct: Your answer is correct.

2

-2.81250

Correct: Your answer is correct.

3

-3.04688

Correct: Your answer is correct.

4

-2.98828

Correct: Your answer is correct.

5

-3.00293

Correct: Your answer is correct.

6

-2.99927

Correct: Your answer is correct.

7

-3.00018

Correct: Your answer is correct.

8

-2.99995

Correct: Your answer is correct.

9

-3.00001

Correct: Your answer is correct.

10

-3.00000

Correct: Your answer is correct.

Graph both the sequence of terms and the sequence of partial sums on the same screen.

WebAssign Plot WebAssign Plot

WebAssign Plot WebAssign Plot

Correct: Your answer is correct. (The one converging near -3, black dots)

Is the series convergent or divergent?

convergent

Correct: Your answer is correct.

If it is convergent, find the sum. (If the quantity diverges, enter DIVERGES.)

set up calculations to determine convergence (geometric)

a/1-r

a=15/-4 , r=1/-4

-3

Correct: Your answer is correct.

Answer 2

Given


`\Sigma_(n=1)^\infty15(-4)^n`

The first 10 partial sums are as follows:


`S_1=\Sigma_(n=1)^(1)15(-4)^n=15(-4)=\bold{-60} \\ \\ S_2=\Sigma_(n=1)^(2)15(-4)^n=\Sigma_(n=1)^(1)15(-4)^n+15(-4)^2 \\ =-60+15(16)=-60+240=\bold{180} \\ \\ S_3=\Sigma_(n=1)^(3)15(-4)^n=\Sigma_(n=1)^(2)15(-4)^n+15(-4)^3 \\ =180+15(-64)=180-960=\bold{-780} \\ \\ S_4=\Sigma_(n=1)^(4)15(-4)^n=\Sigma_(n=1)^(3)15(-4)^n+15(-4)^4 \\ =-780+15(256)=-780+3,840=\bold{3,060} \\ \\ S_5=\Sigma_(n=1)^(5)15(-4)^n=\Sigma_(n=1)^(4)15(-4)^n+15(-4)^5 \\ =3,060+15(-1,024)=3,060-15,360=\bold{-12,300}`


`S_6=\Sigma_(n=1)^(6)15(-4)^n=\Sigma_(n=1)^(5)15(-4)^n+15(-4)^6 \\ =-12,300+15(4,096)=-12,300+61,440=\bold{49,140}`

The rest of the partial sums can be obtained in similar way.