17. Let $a_{k}$ be the coefficient of $x^{k}$ in the expansion of $(1+2 x)^{100}$, where $0 \leq k \leq 100$. Find the number of integers $r: 0 \leq r \leq 99$ such that $a_{r}<a_{r+1}$.
17. Answer: 67 .
Let $a_{k}$ be the coefficient of $x^{k}$ in the expansion of $(1+2 x)^{100}$. Then
$$
a_{k}=2^{k}\binom{100}{k} .
$$

Thus
$$
\frac{a_{r+1}}{a_{r}}=2\binom{100}{r+1} /\binom{100}{r}=\frac{2(100-r)}{r+1} .
$$

Solving the inequality
$$
\frac{2(100-r)}{r+1}>0
$$
gives the solution $r<199 / 3$. It implies that
$$
a_{0}<a_{1}<\cdots<a_{66}<a_{67} \geq a_{68} \geq \cdots \geq a_{100} \text {. }
$$

Hence the answer is 67 .