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 + ...
Accepted Solution
A:
Answer:
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.