Sequences & Series
Closed-Form Sum of Linear Recurrence Sequence
nta_pyq_2025_apr
Grade 11
Question:
Let $a_0=0$, $a_1=\dfrac{1}{2}$, and $2a_{n+2}=5a_{n+1}-3a_n$ for $n\geq0$. Then $\displaystyle\sum_{k=1}^{100}a_k$ equals
$3a_{99}-100$
$3a_{100}-100$
$3a_{99}+100$
$3a_{100}+100$
Step-by-Step Solution
Key Concept: Solve the recurrence to get $a_n=(3/2)^n-1$, then compute the sum as a geometric series minus 100; rewrite the GP partial sum in terms of $a_{100}$.
Characteristic equation: $2r^2-5r+3=0\Rightarrow(2r-3)(r-1)=0\Rightarrow r=\frac{3}{2}$ or $1$.
$a_n=A+B(3/2)^n$. Initial conditions: $a_0=0\Rightarrow A+B=0$; $a_1=1/2\Rightarrow A+3B/2=1/2$. Solving: $B=1$, $A=-1$.
$$a_n=\left(\frac{3}{2}\right)^n-1.$$
$$\sum_{k=1}^{100}a_k=\sum_{k=1}^{100}\left(\frac{3}{2}\right)^k-100=\frac{3/2\cdot((3/2)^{100}-1)}{1/2}-100=3\left(\left(\frac{3}{2}\right)^{100}-1\right)-100.$$
Since $a_{100}=(3/2)^{100}-1$, $(3/2)^{100}=a_{100}+1$:
$$=3a_{100}+3-3-100=3a_{100}-100.$$
Correct Answer: 2