Let a_1=b_1=1, and for n2: a_n = a_n-1+(n-1), b_n = b_n-1+a_n-1.

Question

Let $a_1=b_1=1$, and for $n\ge2$: $a_n = a_{n-1}+(n-1)$, $b_n = b_{n-1}+a_{n-1}$. If $S=\displaystyle\sum_{n=1}^{10}\frac{b_n}{2^n}$ and $T=\displaystyle\sum_{n=1}^{8}\frac{n}{2^{n-1}}$, then $2^7(2S-T)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 461

Accepted Answer

Here $a_n = 1+\dfrac{(n-1)n}{2}$ (so $a_1,\ldots,a_9 = 1,2,4,7,11,16,22,29,37$) and $b_n$ accumulates these. Manipulating $S$ and $\tfrac{S}{2}$ (a standard arithmetico-geometric telescoping) and comparing with $T$, using $b_{10}-b_1 = a_1+a_2+\cdots+a_9 = 130$ (so $b_{10}=130$), one obtains

$$2^7(2S-T) = 2^8(a_1+b_1) - \frac{(b_{10}+2a_9)}{4} = 2^8(2) - \frac{(130+74)}{4} = 461.$$

Solutions · 1