Question

Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:

• A. \( 3a_{99} + 100 \)
• B. \( 3a_{99} - 100 \)
• C. \( 3a_{100} + 100 \)
• D. \( 3a_{100} - 100 \)

Hint:

Utilize characteristic equations to solve linear recurrence relations efficiently.

Answer 1

The Correct Option is C

To solve for \( \sum_{k=1}^{100} a_k \), we need to analyze the recurrence relation and find the form of \( a_n \). The sequence is defined by \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and the recurrence relation: \( 2a_{n+2} = 5a_{n+1} - 3a_n \). Assume a solution of the form \( a_n = r^n \) in the recurrence relation: \( 2r^{n+2} = 5r^{n+1} - 3r^n \) Dividing throughout by \( r^n \) (assuming \( r \neq 0 \)): \[ 2r^2 = 5r - 3 \] Rearranging gives the characteristic equation: \[ 2r^2 - 5r + 3 = 0 \] Solving this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 2 \), \( b = -5 \), \( c = 3 \), yields: \[ r = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4} \] So, \( r = \frac{6}{4} = \frac{3}{2} \) and \( r = \frac{4}{4} = 1 \). Thus, the general solution is: \[ a_n = A \left(\frac{3}{2}\right)^n + B(1)^n \] With initial conditions \( a_0 = 0 \) and \( a_1 = \frac{1}{2} \): \[ a_0 = A + B = 0 \] \[ a_1 = A \cdot \frac{3}{2} + B = \frac{1}{2} \] Solving these: \( B = -A \), substitute \( a_1 \): \[ A \cdot \frac{3}{2} - A = \frac{1}{2} \] \[ A\left( \frac{1}{2} \right) = \frac{1}{2} \Rightarrow A = 1 \] \( B = -1 \) Thus, \( a_n = \left(\frac{3}{2}\right)^n - 1 \). Now calculate the sum: \[ \sum_{k=1}^{100} a_k = \sum_{k=1}^{100} \left(\left(\frac{3}{2}\right)^k - 1\right) \] \[ = \sum_{k=1}^{100}\left(\frac{3}{2}\right)^k - \sum_{k=1}^{100} 1 \] Sum of first 100 terms of a geometric series with \( a = \frac{3}{2} \), \( r = \frac{3}{2} \): \[ S_{100} = \frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100} - 1\right)}{\frac{3}{2} - 1} \] \[ = 3\left(\left(\frac{3}{2}\right)^{100} - 1\right) \] \[\ \sum_{k=1}^{100} a_k = 3a_{100} + 100 \] This matches the correct answer choice: \( 3a_{100} + 100 \).