Competition Math - Algebra

The product can be rewritten as:

$\begin{aligned} P & = \prod_{k=2}^n \left(1-\frac 1{k^2} \right) = \prod_{k=2}^n \frac {(k-1)(k+1)}{k^2} = \frac {(n-1)!(n+1)!}{2!(n!)^2} = \frac {n+1}{2n} = \frac {n+1}{16} \end{aligned}$

Therefore $n = \boxed 8$ .

We can change (1 - $\frac{1}{2^2}$ ) to ( $\frac{2^2}{2^2}$ - $\frac{1}{2^2}$ ), which simplifies to $\frac{2^2 - 1}{2^2}$ . Simplifying the rest of the equation, we get ( $\frac{2^2 - 1}{2^2}$ ) ( $\frac{3^2 - 1}{3^2}$ ) ... ( $\frac{n^2 - 1}{n^2}$ ) = $\frac{n + 1}{16}$ . Since we know that $a^2$ - $b^2$ = ( $a$ - $b$ ) ( $a$ + $b$ ), we can simplify ( $2^2 - 1$ ) to (2 - 1) (2 + 1). Doing the same thing to the rest of the equation, we get ( $\frac{( 2 - 1 ) ( 2 + 1 )}{2^2}$ ) ( $\frac{( 3 - 1 ) ( 3 + 1 )}{3^2}$ ) ... ( $\frac{( n - 1 ) ( n + 1 )}{n^2}$ ) = $\frac{n + 1}{16}$ . Simplifying that, we get $\frac{( 2 - 1 ) ( 2 + 1 ) ( 3 - 1 ) ( 3 + 1 ) ... ( n - 1 ) ( n + 1 )}{2^2 3^3... n^2}$ = $\frac{n + 1}{16}$ . Focusing on the numerator, $( 2 - 1 ) ( 2 + 1 ) ( 3 - 1 ) ( 3 + 1 ) ( 4 - 1 ) ( 4 + 1 ) ( 5 - 1 ) ( 5 + 1 ) ( 6 - 1 ) ( 6 + 1 ) ... ( n - 1 ) ( n + 1 )$ , we can see that (2 + 1) and (4 - 1) both equal 3, (3 + 1) and (5 - 1) both equal 4, (4 + 1) and (6 - 1) both equal 5, and so on. Rearranging the numerator, we get $( 2 - 1 ) ( 3 - 1 ) ( 2 + 1 ) ( 4 - 1 ) ( 3 + 1 ) ( 5 - 1 ) ( 4 + 1 ) ( 6 - 1 ) ( 5 + 1 ) ( 7 - 1 ) ( 6 + 1 ) ( 8 - 1 ) ... ( (n - 3) + 1 ) ( (n - 1) - 1 ) ( (n - 2) + 1 ) ( n - 1 ) ( (n - 1) + 1 ) ( n + 1 )$ . Simplifying, we get $1 × 2 × 3^2 4^2 5^2 6^2 7^2 ... (n-2)^2 (n-1)^2 × n × (n+1)$ . Adding this to the equation, we get $\frac{1 × 2 × 3^2 4^2 5^2 6^2 7^2 ... (n-2)^2 (n-1)^2 × n × (n+1)}{2^2 3^3... n^2}$ = $\frac{n + 1}{16}$ . Now, we can simplify the same factors in the numerator and denominator to get $\frac{n+1}{2n}$ = $\frac{n + 1}{16}$ . Since the numerators are the same in both sides of the equation, we know 2 $n$ = 16. Dividing by two on both sides, we find the solution of $n = \boxed{8}$ .

Using the differences of two squares, the equation can be expressed as (1- $\frac{1}{2}$ )(1+ $\frac{1}{2}$ )(1- $\frac{1}{3}$ )(1+ $\frac{1}{3}$ )...(1- $\frac{1}{n}$ )(1+ $\frac{1}{n}$ ).

After simplifying, $\frac{3}{2}$ $\times$ $\frac{1}{2}$ $\times$ $\frac{4}{3}$ $\times$ $\frac{2}{3}$ $\times$ ... $\times$ $\frac{n+1}{n}$ $\times$ $\frac{n-1}{n}$

And if we rearrange it such that the $\frac{n-1}{n}$ portion comes before the $\frac{n+1}{n}$ portion of (1- $\frac{1}{n^2}$ ), $\frac{1}{2}$ $\times$ $\frac{3}{2}$ $\times$ $\frac{2}{3}$ $\times$ $\frac{4}{3}$ $\times$ ... $\times$ $\frac{n-1}{n}$ $\times$ $\frac{n+1}{n}$

The $\frac{n+1}{n}$ portion of the kth term when multiplied with the $\frac{n-1}{n}$ portion of the (k+1)th term equals to one: ( $\frac{k+1}{k}$ ) $\times$ $\frac{(k+1)-1}{(k+1)}$ = 1

Therefore, everything in the middle gets cut out. What's left is the $\frac{n-1}{n}$ portion of the very first term (1- $\frac{1}{2}$ ) and the $\frac{n+1}{n}$ portion of the nth term. Therefore, leaving you with $\frac{1}{2}$ $\times$ $\frac{n+1}{n}$ = $\frac{n+1}{2n}$ => $\frac{n+1}{2n}$ = $\frac{n+1}{16}$

Thus, for the equation to be true, n = $\boxed{8}$ .

$\begin{aligned} \frac{n+1}{16} &= \bigg( 1- \frac{1}{2^2}\bigg) \bigg( 1- \frac{1}{3^2}\bigg) \bigg( 1- \frac{1}{4^2}\bigg) \cdots \bigg( 1- \frac{1}{n^2}\bigg) \\ \\ &= \bigg(\frac{(2-1)(2+1)}{2^2}\bigg) \bigg(\frac{(3-1)(3+1)}{3^2}\bigg) \bigg(\frac{(4-1)(4+1)}{4^2}\bigg) \cdots \bigg(\frac{(n-1)(n+1)}{n^2}\bigg) \\ \\ &= \frac{{\color{#D61F06}{(1)}}\bcancel{(3)}}{{\color{#D61F06}{(2)}}\cancel{(2)}} \frac{\cancel{(2)}\bcancel{(4)}}{\bcancel{(3)}\cancel{(3)}} \frac{\cancel{(3)}\bcancel{(5)}}{\bcancel{(4)}\cancel{(4)}} \cdots \bigg(\frac{\cancel{(n-1)}{\color{#D61F06}{(n+1)}}}{\bcancel{(n)}{\color{#D61F06}{(n)}}}\bigg) \\ \\ &= \frac{1}{2} \cdot \frac{n+1}{n} \implies \frac{n+1}{2n} \\ \\ \frac{n+1}{16} &= \frac{n+1}{2n} \\ \\ 2n &= 16 \implies n = \boxed{8} \end{aligned}$

From python

def f(x):

z=1

for a in range(2,x+1):

    l=1-(1/(a**2))

    z*=l   

return z

for q in range(1,100):

if f(q)==(1+q)/16:

    print(q)

Output:8