Question

Show work and explain with formulas.

6. Given the series: 4 + 12 + 36 + ...

find the 11th term of the series and find the sum of the first 11 terms.

7. A ball is dropped from a height of 40 feet and bounces back up to 90% of its previous height on each successive bounce. How far will the ball have traveled by the time it comes to a stop?

8. For each series below, decide whether it converges or diverges. Find the sum, if it exists.

a) 80 + 20 + 5 + 5/4 + ...
b) 2/9 + 4/3 + 8 + ...​

Answer 1

Q6. a₁₁ = 236 196; S₁₁ = 354 292

Q7. 354 292

Q8. a) Converges, S = 320/3; b) diverges

Step-by-step explanation:

6. Geometric sequence

The first three terms of your sequence are 4, 12, 36.

Each term differs from the previous one by a factor of 3, so it is a geometric sequence.

Each term has the form

aₙ = a₁rⁿ⁻¹

In your sequence, a₁ = 4 and r = 3.

Thus, the formula for the nth term is

aₙ = 4(3)ⁿ⁻¹

The 11th term is

a₁₁ = 4(3)¹¹⁻¹ = 4(3)¹⁰ = 4 × 59 049 = 236 196

The formula for the sum of the first n terms of a geometric series is

Sum = a₁[(1 - rⁿ)/(1 - r)]

For the sum over the first 11 terms,

Sum = 4[(1 - 3¹¹)/(1 - 3)

= 4(1 - 177 147)/(-2)

= -2(-177 146)

= 354 292

7. Bouncing ball

Mathematically, the ball never stops bouncing. The height of each bounce just gets infinitesimally small.

So , you have an infinite geometric series in which the first term is 40 ft and each successive term is 90 % of the previous term.

The general formula for the nth term is

aₙ = a₁rⁿ⁻¹

with a₁ = 40 ft and r = 0.90.

Since |r| <1, we have a convergent series, and the formula for the sum is

S = a₁/(1-r)

∴ S = 40/(1 - 0.90) = 40/0.10 = 400 ft

The bouncing ball will have travelled 400 ft.

8. Test for convergence

a) 80 + 20 + 5 + 5/4 + ...

r = a₂/a₁ = 20/80 = ¼

r < 1, so the series converges.

S = 80/( 1 - ¼) = 80/¾ = 80 × ⁴/₃ = 320/3

The sum of the series is 320/3.

b) 2/9 + 4/3 + 8 + ...

r = a₂/a₁ = (⁴/₃)/(²/₉) = ⁴/₃ × ⁹/₂ = 2 × 3 = 6.

r > 1, so the series diverges.

Answer 2

6 Answer: 236,196 and 354,294

Explanation:

`\text{Find the 11th term:}\\a_1=4\qquad r=(12)/(4)=3\qquad n=11\\\\a_n=a_1\cdot r^(n-1)\\a_(11)=4\cdot (3)^(11-1)\\.\quad =4\cdot3^(10)\\.\quad =4\cdot 59,049\\.\quad =\large\boxed{236,196}\\\\\\\text{Find the sum:}\\S_n=(a_1(1-r^n))/(1-r)\\\\\\S_(11)=(4(1-3^(11)))/(1-3)\\\\\\.\quad =(4(1-177,147))/(-2)\\\\.\quad =-2(-177,146)\\\\.\quad =\large\boxed{354,294}`

7 Answer: 400

Explanation:

`\sum\limits^(n=1)_\infty 40(0.9)^(n-1)\\\\a_1=40\qquad r=0.9\\\\S_\infty =(a_1)/(1-r)\\\\\\.\quad =(40)/(1-0.9)\\\\\\.\quad =(40)/(0.1)\\\\\\.\quad =\large\boxed{400}`

8 Answer: a) converges

b) diverges

Explanation:

A series converges if: -1 < r < 1

a) 80 + 20 + 5 +
`(5)/(4)`

`a)\ r=(a_2)/(a_1)=(20)/(80)=(1)/(4)\implies \text{converges}\\\\\\b)\ r={a_2}/{a_1}=(4)/(3)/ (2)/(9)=(4)/(3)* (9)/(2)=2* 3=6\implies \text{diverges}`