Sequences & Series
Recurrence sequences and series manipulation
nta_pyq_2023_jan
Grade 11
Question:
Let $a_1 = b_1 = 1$ and $a_n = a_{n-1} + (n-1)$, $b_n = b_{n-1} + a_{n-1}$, $\forall n \geq 2$. If $S = \displaystyle\sum_{n=1}^{10} \dfrac{b_n}{2^n}$ and $T = \displaystyle\sum_{n=1}^{8} \dfrac{n}{2^{n-1}}$, then $2^7(2S - T)$ is equal to ______.
Step-by-Step Solution
Key Concept: Find $a_n$ from the recurrence: $a_n = 1 + \frac{(n-1)(n-2)}{2} + (n-1) = \frac{n^2-n+2}{2}$. Then compute $b_{10}$ and $a_9$ using the telescoping, and apply the derived formula $2^7(2S-T) = 2^8(a_1+b_1) - \frac{b_{10}+2a_9}{4}$.
$a_9 = 37$, $b_{10} = 130$. $2^7(2S-T) = 2^8(1+1) - \frac{130+74}{4} = 512 - \ frac{204}{4} = 512 - 51 = 461$.
Correct Answer: 461