IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Discover comprehensive answers to your questions from our community of knowledgeable experts.
Sagot :
Sure! Let's break down the problem and find the solution step by step.
### Part (i): Finding the sum [tex]\( S \)[/tex] of the first [tex]\( n \)[/tex] terms of the sequence whose [tex]\( r \)[/tex]-th term is [tex]\( 4 \cdot 2^r \)[/tex]
The sequence given is:
[tex]\[ a_r = 4 \cdot 2^r \][/tex]
We recognize that this sequence is a geometric sequence where each term is multiplied by a common ratio. To write it in a standard form, consider the first few terms:
- When [tex]\( r = 1 \)[/tex],
[tex]\[ a_1 = 4 \cdot 2^1 = 8 \][/tex]
- When [tex]\( r = 2 \)[/tex],
[tex]\[ a_2 = 4 \cdot 2^2 = 16 \][/tex]
- When [tex]\( r = 3 \)[/tex],
[tex]\[ a_3 = 4 \cdot 2^3 = 32 \][/tex]
Thus, the general term can be written as:
[tex]\[ a_r = 4 \cdot 2^{r-1} \][/tex]
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric sequence is given by:
[tex]\[ S_n = a_1 \cdot \frac{r^n - 1}{r - 1} \][/tex]
In our sequence:
- The first term [tex]\( a_1 = 8 \)[/tex] (since [tex]\( 4 \cdot 2^1 \)[/tex])
- The common ratio [tex]\( r = 2 \)[/tex]
Substituting these values into the formula, we get:
[tex]\[ S_n = 8 \cdot \frac{2^n - 1}{2 - 1} = 8 \cdot (2^n - 1) \][/tex]
For example, let's take [tex]\( n = 10 \)[/tex]:
[tex]\[ S_{10} = 8 \cdot (2^{10} - 1) \][/tex]
[tex]\[ S_{10} = 8 \cdot (1024 - 1) \][/tex]
[tex]\[ S_{10} = 8 \cdot 1023 \][/tex]
[tex]\[ S_{10} = 8184 \][/tex]
### Part (ii): Finding the value of [tex]\( n \)[/tex] for which the difference between [tex]\( S \)[/tex] and 4 is less than 104
We need to find [tex]\( n \)[/tex] such that:
[tex]\[ |S_n - 4| < 104 \][/tex]
From Part (i), we have:
[tex]\[ S_n = 8 \cdot (2^n - 1) \][/tex]
We want the difference [tex]\( |S_n - 4| \)[/tex] to be less than 104:
[tex]\[ |8 \cdot (2^n - 1) - 4| < 104 \][/tex]
Let's simplify the expression inside the absolute value:
[tex]\[ 8 \cdot (2^n - 1) - 4 = 8 \cdot 2^n - 8 - 4 = 8 \cdot 2^n - 12 \][/tex]
Thus, the condition becomes:
[tex]\[ |8 \cdot 2^n - 12| < 104 \][/tex]
Solving for the inequality:
[tex]\[ 8 \cdot 2^n - 12 < 104 \][/tex]
[tex]\[ 8 \cdot 2^n - 12 > -104 \][/tex]
First inequality:
[tex]\[ 8 \cdot 2^n < 116 \][/tex]
[tex]\[ 2^n < 14.5 \][/tex]
Taking the base 2 logarithm:
[tex]\[ n < \log_2 (14.5) \approx 3.857 \][/tex]
Since [tex]\( n \)[/tex] must be an integer:
[tex]\[ n \leq 3 \][/tex]
Second inequality:
[tex]\[ 8 \cdot 2^n > -92 \][/tex]
[tex]\[ 2^n > -11.5 \][/tex]
Since [tex]\( 2^n \)[/tex] is always positive, this inequality is always true.
Therefore, the integer value [tex]\( n = 1 \)[/tex] satisfies the condition where the difference between [tex]\( S_n \)[/tex] and 4 is less than 104.
This means the value of [tex]\( n \)[/tex] that satisfies the condition is [tex]\( 1 \)[/tex].
### Summary:
(i) The sum [tex]\( S \)[/tex] of the first 10 terms of the sequence is [tex]\( 4092 \)[/tex].
(ii) The value [tex]\( n \)[/tex] for which the difference between [tex]\( S \)[/tex] and 4 is less than 104 is [tex]\( 1 \)[/tex].
### Part (i): Finding the sum [tex]\( S \)[/tex] of the first [tex]\( n \)[/tex] terms of the sequence whose [tex]\( r \)[/tex]-th term is [tex]\( 4 \cdot 2^r \)[/tex]
The sequence given is:
[tex]\[ a_r = 4 \cdot 2^r \][/tex]
We recognize that this sequence is a geometric sequence where each term is multiplied by a common ratio. To write it in a standard form, consider the first few terms:
- When [tex]\( r = 1 \)[/tex],
[tex]\[ a_1 = 4 \cdot 2^1 = 8 \][/tex]
- When [tex]\( r = 2 \)[/tex],
[tex]\[ a_2 = 4 \cdot 2^2 = 16 \][/tex]
- When [tex]\( r = 3 \)[/tex],
[tex]\[ a_3 = 4 \cdot 2^3 = 32 \][/tex]
Thus, the general term can be written as:
[tex]\[ a_r = 4 \cdot 2^{r-1} \][/tex]
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a geometric sequence is given by:
[tex]\[ S_n = a_1 \cdot \frac{r^n - 1}{r - 1} \][/tex]
In our sequence:
- The first term [tex]\( a_1 = 8 \)[/tex] (since [tex]\( 4 \cdot 2^1 \)[/tex])
- The common ratio [tex]\( r = 2 \)[/tex]
Substituting these values into the formula, we get:
[tex]\[ S_n = 8 \cdot \frac{2^n - 1}{2 - 1} = 8 \cdot (2^n - 1) \][/tex]
For example, let's take [tex]\( n = 10 \)[/tex]:
[tex]\[ S_{10} = 8 \cdot (2^{10} - 1) \][/tex]
[tex]\[ S_{10} = 8 \cdot (1024 - 1) \][/tex]
[tex]\[ S_{10} = 8 \cdot 1023 \][/tex]
[tex]\[ S_{10} = 8184 \][/tex]
### Part (ii): Finding the value of [tex]\( n \)[/tex] for which the difference between [tex]\( S \)[/tex] and 4 is less than 104
We need to find [tex]\( n \)[/tex] such that:
[tex]\[ |S_n - 4| < 104 \][/tex]
From Part (i), we have:
[tex]\[ S_n = 8 \cdot (2^n - 1) \][/tex]
We want the difference [tex]\( |S_n - 4| \)[/tex] to be less than 104:
[tex]\[ |8 \cdot (2^n - 1) - 4| < 104 \][/tex]
Let's simplify the expression inside the absolute value:
[tex]\[ 8 \cdot (2^n - 1) - 4 = 8 \cdot 2^n - 8 - 4 = 8 \cdot 2^n - 12 \][/tex]
Thus, the condition becomes:
[tex]\[ |8 \cdot 2^n - 12| < 104 \][/tex]
Solving for the inequality:
[tex]\[ 8 \cdot 2^n - 12 < 104 \][/tex]
[tex]\[ 8 \cdot 2^n - 12 > -104 \][/tex]
First inequality:
[tex]\[ 8 \cdot 2^n < 116 \][/tex]
[tex]\[ 2^n < 14.5 \][/tex]
Taking the base 2 logarithm:
[tex]\[ n < \log_2 (14.5) \approx 3.857 \][/tex]
Since [tex]\( n \)[/tex] must be an integer:
[tex]\[ n \leq 3 \][/tex]
Second inequality:
[tex]\[ 8 \cdot 2^n > -92 \][/tex]
[tex]\[ 2^n > -11.5 \][/tex]
Since [tex]\( 2^n \)[/tex] is always positive, this inequality is always true.
Therefore, the integer value [tex]\( n = 1 \)[/tex] satisfies the condition where the difference between [tex]\( S_n \)[/tex] and 4 is less than 104.
This means the value of [tex]\( n \)[/tex] that satisfies the condition is [tex]\( 1 \)[/tex].
### Summary:
(i) The sum [tex]\( S \)[/tex] of the first 10 terms of the sequence is [tex]\( 4092 \)[/tex].
(ii) The value [tex]\( n \)[/tex] for which the difference between [tex]\( S \)[/tex] and 4 is less than 104 is [tex]\( 1 \)[/tex].
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Discover insightful answers at IDNLearn.com. We appreciate your visit and look forward to assisting you again.
