5. Fer which values of $$ k $$ will the series, $$ 4\left(\frac{1-k}{5}\right)+8\left(\frac{1-k}{5}\right)^{2}+16\left(\frac{1-k}{5}\right)^{3} $$ converge? 6. (4) Calculate the value of $$ \sum_{k=1}^{50}(100-3 k) $$ a) b) Calculate the value of $$ \sum_{n=3}^{6}\left(2 n^{2}-1\right) $$ 7. The Ith term of an arithmetic sequence is 32 , and the $$ s $$ ( 18$$ ) $$ of the first 11 terms is 187 . Calculate the 21 st term. 8. Determine the value of $$ m, \sum_{p=1}^{m}(-8)\left(\frac{1}{2}\right)^{p-1}=-15 \frac{3}{4} $$ (

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Let's solve each of the problems step by step. ### Problem 5: Convergence of the Series The series given is: $$ 4\left(\frac{1-k}{5}\right) + 8\left(\frac{1-k}{5}\right)^{2} + 16\left(\frac{1-k}{5}\right)^{3} $$ This can be recognized as a geometric series where the first term $$ a = 4\left(\frac{1-k}{5}\right) $$ and the common ratio $$ r = \frac{1-k}{5} $$. For a geometric series to converge, the absolute value of the common ratio must be less than 1: $$ \left|\frac{1-k}{5}\right| < 1 $$ Now, let's solve this inequality: 1. Multiply both sides by 5 (since 5 is positive, the inequality remains the same): $$ |1-k| < 5 $$ 2. This can be split into two inequalities: $$ -5 < 1-k < 5 $$ 3. Rearranging gives: $$ -4 < -k < 4 $$ 4. Multiplying through by -1 (which reverses the inequalities): $$ 4 > k > -4 $$ Thus, the series converges for: $$ -4 < k < 4 $$ ### Problem 6: Summation Calculations #### Part a: Calculate $$ \sum_{k=1}^{50}(100-3k) $$ This is an arithmetic series where: - The first term $$ a = 100 - 3(1) = 97 $$ - The last term $$ l = 100 - 3(50) = -50 $$ - The number of terms $$ n = 50 $$ The sum of an arithmetic series can be calculated using the formula: $$ S_n = \frac{n}{2} (a + l) $$ Substituting the values: $$ S_{50} = \frac{50}{2} (97 - 50) = 25 \times 47 = 1175 $$ #### Part b: Calculate $$ \sum_{n=3}^{6}(2n^{2}-1) $$ Calculating each term from $$ n = 3 $$ to $$ n = 6 $$: - For $$ n = 3 $$: $$ 2(3^2) - 1 = 2(9) - 1 = 18 - 1 = 17 $$ - For $$ n = 4 $$: $$ 2(4^2) - 1 = 2(16) - 1 = 32 - 1 = 31 $$ - For $$ n = 5 $$: $$ 2(5^2) - 1 = 2(25) - 1 = 50 - 1 = 49 $$ - For $$ n = 6 $$: $$ 2(6^2) - 1 = 2(36) - 1 = 72 - 1 = 71 $$ Now summing these values: $$ 17 + 31 + 49 + 71 = 168 $$ ### Problem 7: Arithmetic Sequence Given: - The $$ I $$th term $$ a_{11} = 32 $$ - The sum of the first 11 terms $$ S_{11} = 187 $$ The $$ n $$th term of an arithmetic sequence is given by: $$ a_n = a + (n-1)d $$ The sum of the first $$ n $$ terms is given by: $$ S_n = \frac{n}{2} (2a + (n-1)d) $$ From $$ a_{11} = 32 $$: $$ a + 10d = 32 \quad (1) $$ From $$ S_{11} = 187 $$: $$ \frac{11}{2} (2a + 10d) = 187 $$ Multiplying through by 2: $$ 11(2a + 10d) = 374 $$ Dividing by 11: $$ 2a + 10d = 34 \quad (2) $$ Now we have a system of equations: 1. $$ a + 10d = 32 $$ 2. $$ 2a + 10d = 34 $$ Subtracting (1) from (2): $$ (2a + 10d) - (a + 10d) = 34 - 32 $$ This simplifies to: $$ a = 2 $$ Substituting $$ a = 2 $$ back into (1): $$ 2 + 10d = 32 \implies 10d = 30 \implies d = 3 $$ Now, to find the 21st term: $$ a_{21} = a + 20d = 2 + 20(3) = 2 + 60 = 62 $$ ### Problem 8: Summation Equation We need to solve: $$ \sum_{p=1}^{m}(-8)\left(\frac{1}{2}\right)^{p-1} = -15 \frac{3}{4} $$ This is a geometric series where: - First term $$ a = -8 $$ - Common ratio $$ r = \frac{1}{2} $$ - Number of terms $$ m $$ The sum of the first $$ m $$ terms of a geometric series is given by: $$ S_m = a \frac{1 - r^m}{1 - r} $$ Substituting the values: $$ S_m = -8 \frac{1 - \left(\frac{1}{2}\right)^{m}}{1 - \frac{1}{2}} = -8 \cdot 2 (1 - \left(\frac{1}{2}\right)^{m}) = -16 (1 - \left(\frac{1}{2}\right)^{m}) $$ Setting this equal to $$ -15 \frac{3}{4} = -\frac{63}{4} $$: $$ -16 (1 - \left(\frac{1}{2}\right)^{m}) = -\frac{63}{4} $$ Dividing both sides by -16: $$ 1 - \left(\frac{1}{2}\right)^{m} = \frac{63}{64} $$ Thus: $$ \left(\frac{1}{2}\right)^{m} = 1 - \frac{63}{64} = \frac{1}{64} $$ Taking the logarithm base 2: $$ m = 6 $$ ### Summary of Results 1. The series converges for $$ -4 < k < 4 $$. 2. $$ \sum_{k=1}^{50}(100-3k) = The series converges for $$ -4 < k < 4 $$.

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