Question ID: #438
Let $< a_n >$ be a sequence such that $a_0 = 0$, $a_1 = \frac{1}{2}$ and $2a_{n+2} = 5a_{n+1} – 3a_n$, $n = 0, 1, 2, \dots$. Then $\sum_{k=1}^{100} a_k$ is equal to:
- (1) $3a_{99} – 100$
- (2) $3a_{100} – 100$
- (3) $3a_{100} + 100$
- (4) $3a_{99} + 100$
Solution:
The recurrence relation is $2a_{n+2} – 5a_{n+1} + 3a_n = 0$.
The characteristic equation is $2r^2 – 5r + 3 = 0$.
$$ (2r – 3)(r – 1) = 0 \Rightarrow r = 1, \frac{3}{2} $$
So the general term is $a_n = A(1)^n + B\left(\frac{3}{2}\right)^n = A + B\left(\frac{3}{2}\right)^n$.
Using initial values:
For $n=0$: $A + B = 0 \Rightarrow A = -B$.
For $n=1$: $A + \frac{3}{2}B = \frac{1}{2}$.
Substitute $A = -B$:
$-B + \frac{3}{2}B = \frac{1}{2} \Rightarrow \frac{1}{2}B = \frac{1}{2} \Rightarrow B = 1$.
So $A = -1$.
Thus, $a_n = -1 + \left(\frac{3}{2}\right)^n$.
We need to calculate $S = \sum_{k=1}^{100} a_k$.
$$ S = \sum_{k=1}^{100} \left[ -1 + \left(\frac{3}{2}\right)^k \right] $$
$$ S = \sum_{k=1}^{100} (-1) + \sum_{k=1}^{100} \left(\frac{3}{2}\right)^k $$
$$ S = -100 + \frac{\frac{3}{2}\left( (\frac{3}{2})^{100} – 1 \right)}{\frac{3}{2} – 1} $$
$$ S = -100 + \frac{\frac{3}{2}\left( (\frac{3}{2})^{100} – 1 \right)}{\frac{1}{2}} $$
$$ S = -100 + 3\left[ \left(\frac{3}{2}\right)^{100} – 1 \right] $$
$$ S = -100 + 3\left(\frac{3}{2}\right)^{100} – 3 $$
Now let’s check the options expressed in terms of $a_{n}$.
$a_{100} = -1 + \left(\frac{3}{2}\right)^{100} \Rightarrow \left(\frac{3}{2}\right)^{100} = a_{100} + 1$.
Substitute this into the sum:
$$ S = -100 + 3(a_{100} + 1) – 3 $$
$$ S = -100 + 3a_{100} + 3 – 3 $$
$$ S = 3a_{100} – 100 $$
Ans. (2)
Was this solution helpful?
YesNo