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

Question

10-11-2017
Mathematics

contestada

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

Respuesta :

Otras preguntas

Rank the following elements by electron affinity, from most positive to most negative EA value. Rank from most positive to most negative. To rank items as equiv

To open a 3-digit safe the tip was placed Multiply the first number by 3 and the answer will always be with numbers 1. Multiply the second by 6 and the answer

What is the quotient: (2x2 + 10x + 12) ÷ (x + 3) ? a. 2x + 6 b. 2x – 6 c. 2x + 4 d. 2x – 4, r = 1

If the simple interest on $2000 for 2 years is $320, then what is the interest rate?

ANSWER ASAP 30 POINTS! Keko listed one source of income on his October balance sheet: $10.00 for painting a fence. His total expenses are $33.00. Which two sour

How did the success of West African kingdoms contribute to the development of kingdoms in western and southern Africa? As the populations of western kingdoms g

Marcus is eating a meal that is made up of an entr_e that is 45 calories per ounce and a side dish that is 30 calories per ounce. He wants his meal to total 510

What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The elements have the same number of e

If the following elements were to form ions, they would attain the same number of electrons as which noble gas? elements: Be, Ca, Al, Rb, Se, F, P noble gas: H

I'm trying to solve 2x^2+3x=2

nobillionaireNobley nobillionaireNobley · Answer 1 · 2017-11-21T09:17:37+01:00

Given

[tex]\Sigma_{n=1}^\infty15(-4)^n[/tex]

The first 10 partial sums are as follows:

[tex]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}[/tex]

[tex]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}[/tex]

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

JayDilla JayDilla · Answer 2 · 2020-10-27T17:07:03+01:00

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