Question

Qual é o valor do produto $\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\dots\left(1-\dfrac{1}{225}\right)$ ?

Answer 1

Cada um dos fatores é uma diferença de quadrados, isto é, $a^2-b^2$, em que, $a=1$ e $b=\dfrac{1}{c^2}=\left(\dfrac{1}{c}\right)^2$.

Usando a fatoração $a^2-b^2=(a-b)(a+b)$, obtemos:

$\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\dots\left(1-\dfrac{1}{225}\right)$

$=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{2^4}\right)\dots\left(1-\dfrac{1}{15^2}\right)$

$=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\dots\left(1-\dfrac{1}{15}\right)\left(1+\dfrac{1}{15}\right)$

$=\dfrac{1}{2}\times\underbrace{\dfrac{3}{2}\times\dfrac{2}{3}}_{1}\times\underbrace{\dfrac{4}{3}\times\dfrac{3}{4}}_{1}\times\underbrace{\dfrac{5}{4}\times\dots}_{1}\times\underbrace{\dots\times\dfrac{14}{15}}_{1}\times\dfrac{16}{15}$

$=\dfrac{1}{2}\times\dfrac{16}{15}=\boxed{\dfrac{8}{15}}$.