Answer
**Part 1: Quadratic Sequence**
- \( x = 6.5 \)
- \( y = 44.5 \)
**Part 2: Series Calculation**
- **Sum of the series:** 3553
- **Sigma-notation:** \( \sum_{k=0}^{33} (220 - 7k) \)
**Part 3: Bouncing Ball**
- **Total distance:** 285 m
**Part 4: Series Convergence**
- **Convergence range for \( t \):** \( -4 < t < 2 \)
- **For \( t = 15 \):** The series does not converge.
**Part 5: Sum of the First \( n \) Terms**
- **70th term:** \( T_{70} = 2^{70-1} + 3 = 2^{69} + 3 \)
Solution
Let's break down the problem step by step.
### Part 1: Quadratic Sequence
The sequence given is \( 6, x, 26, 45, y, \ldots \).
1. **Identify the differences:**
- First difference: \( x - 6 \)
- Second difference: \( 26 - x \)
- Third difference: \( 45 - 26 = 19 \)
- Fourth difference: \( y - 45 \)
2. **Calculate the first differences:**
- First difference: \( d_1 = x - 6 \)
- Second difference: \( d_2 = 26 - x \)
- Third difference: \( d_3 = 19 \)
- Fourth difference: \( d_4 = y - 45 \)
3. **Set up the equations based on the differences:**
- The second differences should be constant for a quadratic sequence.
- Thus, \( d_2 - d_1 = d_3 \) and \( d_3 - d_2 = d_4 \).
4. **Solve for \( x \) and \( y \):**
- From \( d_2 - d_1 = 19 \):
\[
(26 - x) - (x - 6) = 19
\]
\[
32 - 2x = 19 \implies 2x = 13 \implies x = 6.5
\]
- From \( d_3 - d_2 = d_4 \):
\[
19 - (26 - x) = y - 45
\]
Substituting \( x = 6.5 \):
\[
19 - (26 - 6.5) = y - 45
\]
\[
19 - 19.5 = y - 45 \implies -0.5 = y - 45 \implies y = 44.5
\]
Thus, the values are \( x = 6.5 \) and \( y = 44.5 \).
### Part 2: Series Calculation
The series is \( 220 + 213 + 206 + \ldots - 11 \).
#### 2.2.1 Calculate the sum of the series.
1. **Identify the first term and common difference:**
- First term \( a = 220 \)
- Common difference \( d = 213 - 220 = -7 \)
2. **Find the number of terms:**
- The last term is \( -11 \).
- Use the formula for the \( n \)-th term of an arithmetic series:
\[
a_n = a + (n-1)d
\]
Setting \( a_n = -11 \):
\[
-11 = 220 + (n-1)(-7)
\]
\[
-11 - 220 = -7(n-1) \implies -231 = -7(n-1) \implies n-1 = 33 \implies n = 34
\]
3. **Calculate the sum:**
- Use the sum formula for an arithmetic series:
\[
S_n = \frac{n}{2} (a + a_n)
\]
\[
S_{34} = \frac{34}{2} (220 - 11) = 17 \times 209 = 3553
\]
#### 2.2.2 Write the series in sigma-notation.
The series can be expressed in sigma notation as:
\[
\sum_{k=0}^{33} (220 - 7k)
\]
### Part 3: Bouncing Ball
A ball is dropped from a height of \( 15 \) m and loses \( 10\% \) of its height on each bounce.
1. **Calculate the total distance:**
- The height after each bounce is \( 90\% \) of the previous height.
- The heights form a geometric series:
- First drop: \( 15 \)
- First bounce: \( 15 \times 0.9 = 13.5 \)
- Second drop: \( 13.5 \)
- Second bounce: \( 13.5 \times 0.9 = 12.15 \)
- And so on...
2. **Total distance:**
- The total distance \( D \) is:
\[
D = 15 + 2(13.5 + 12.15 + \ldots)
\]
- The series \( 13.5 + 12.15 + \ldots \) is a geometric series with first term \( a = 13.5 \) and common ratio \( r = 0.9 \).
- The sum of an infinite geometric series is:
\[
S = \frac{a}{1 - r} = \frac{13.5}{1 - 0.9} = \frac{13.5}{0.1} = 135
\]
- Thus, the total distance:
\[
D = 15 + 2 \times 135 = 15 + 270 = 285
\]
The total distance the ball will bounce cannot exceed \( 290 \) m.
### Part 4: Series Convergence
Given:
\[
25\left(\frac{1-t}{3}\right) + 5\left(\frac{1-1}{3}\right)^{2} + \left(\frac{1-1}{3}\right)^{3} + \ldots
\]
#### 2.4.1 For which value(s) of \( t \) will the series converge?
The series converges if the common ratio \( r < 1 \):
\[
\left|\frac{1-t}{3}\right| < 1
\]
This leads to:
\[
-1 < \frac{1-t}{3} < 1
\]
Multiplying by \( 3 \):
\[
-3 < 1 - t < 3
\]
This gives:
\[
-2 < -t < 4 \implies -4 < t < 2
\]
#### 2.4.2 If \( t = 15 \), calculate the sum to infinity of the series if it exists.
Since \( t = 15 \) is outside the convergence range, the series does not converge.
### Part 5: Sum of the First \( n \) Terms
Given:
\[
S_{\infty} = 2^{x-1} + 3
\]
To find the \( 70^{th} \) term, we need to express the \( n \)-th term \( T_n \):
\[
T_n = S_n - S_{n-1}
\]
Assuming \( S_n = 2^{n-1}
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