just a quadratic equation

Algebra Level 3

find the sum of all positive integers n n such that

( 1 − 1 2 2 ) ( 1 − 1 3 2 ) ( 1 − 1 4 2 ) ⋯ ( 1 − 1 n 2 ) = 8 n 225 (1-\dfrac{1}{2^2})(1-\dfrac{1}{3^2})(1-\dfrac{1}{4^2}) \cdots (1-\dfrac{1}{n^2})= \dfrac{8n}{225}


The answer is 15.

Diego Perez by Diego Perez , 17, Chile

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

X X
May 12, 2018

( 1 − 1 2 2 ) ( 1 − 1 3 2 ) ( 1 − 1 4 2 ) ⋯ ( 1 − 1 n 2 ) = 1 × 3 2 × 2 × 2 × 4 3 × 3 × 3 × 5 4 × 4 × ⋯ × ( n − 1 ) ( n + 1 ) n × n = n + 1 2 n = 8 n 225 (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2}) \cdots (1-\frac{1}{n^2})=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\cdots\times\frac{(n-1)(n+1)}{n\times n}=\frac{n+1}{2n}=\frac{8n}{225}

Solving and get n = 15 n=15

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...