Question

Halla el residuo cuando $\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \cdots + \binom{\binom{40}{2}}{2}$ se divide entre $1000.$

Find the remainder when $\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \cdots + \binom{\binom{40}{2}}{2}$ is divided by $1000.$

Respuesta

Solución:

Como $\binom{n}{2} = \frac{n(n-1)}{2}$ y $\binom{n}{2} - 1 = \frac{(n+1)(n-2)}{2},$ $\begin{aligned} \binom{\binom{n}{2}}{2} \\ &= \frac{1}{2} \cdot \frac{n(n-1)}{2} \\ &\quad {}\cdot \frac{(n+1)(n-2)}{2} \\ &= \small \frac{(n+1)n(n-1)(n-2)}{8} \\ &= 3\binom{n+1}{4}. \end{aligned}$

Por la identidad del palo de hockey, $\begin{aligned} \sum_{n=3}^{40} 3\binom{n+1}{4} &= 3\sum_{k=4}^{41}\binom{k}{4} \\ &= 3\binom{42}{5} \\ &= 3 \cdot 850668 \\ &= 2552004. \end{aligned}$

El residuo al dividir entre $1000$ es $4.$

Since $\binom{n}{2} = \frac{n(n-1)}{2}$ and $\binom{n}{2} - 1 = \frac{(n+1)(n-2)}{2},$ $\begin{aligned} \binom{\binom{n}{2}}{2} \\ &= \frac{1}{2} \cdot \frac{n(n-1)}{2} \\ &\quad {}\cdot \frac{(n+1)(n-2)}{2} \\ &= \small \frac{(n+1)n(n-1)(n-2)}{8} \\ &= 3\binom{n+1}{4}. \end{aligned}$

By the hockey stick identity, $\begin{aligned} \sum_{n=3}^{40} 3\binom{n+1}{4} &= 3\sum_{k=4}^{41}\binom{k}{4} \\ &= 3\binom{42}{5} \\ &= 3 \cdot 850668 \\ &= 2552004. \end{aligned}$

The remainder upon division by $1000$ is $4.$

2022 AIME II Problema 10

Solución:

El Problema 10 en otros años