The greatest integer less than or equal to the sum of the first 100 terms of the sequence …

Question

The greatest integer less than or equal to the sum of the first $100$ terms of the sequence $\dfrac13, \dfrac59, \dfrac{19}{27}, \dfrac{65}{81}, \ldots$ is equal to \_\_\_\_\_\_\_\_.

Answer: 98

Accepted Answer

The numerators satisfy $a_n = 3^n-2^n$ (check: $3-2=1$, $9-4=5$, $27-8=19$, $81-16=65$), so the $n$th term is $T_n = 1-\left(\dfrac23\right)^n$.

$$\sum_{n=1}^{100}T_n = 100 - \sum_{n=1}^{100}\left(\frac23\right)^n = 100 - 2\left[1-\left(\frac23\right)^{100}\right] = 98 + 2\left(\frac23\right)^{100}.$$

Since $2(2/3)^{100}$ is a tiny positive number, the greatest integer $\le$ this sum is $\boxed{98}$.

Solutions · 1