Trig sum

Algebra Level pending

If a geometric progression has $a_{1}=2.tg (\frac{\pi}{5})$ as its first term and $q=tg^{2} (\frac{\pi}{5})$ as its ratio of progression, the sum of all its terms, for $n$ terms, with $n$ varying from $1$ to $\infty$ , can be expressed as $\zeta=tg (\frac{s.\pi}{r})$ , such that $s$ and $r$ are coprime positive integers. So, what's $s+r$ ?

Details and assumptions

You might desconsider answers out of the interval $[0,\frac{\pi}{2}]$

The answer is 7.

1 solution

Mikael Marcondes
Jan 10, 2015

The sum of the terms of a geometric progression can be written as:

$\sum\limits_{n=1}^{n} a_{1}.q^{(n-1)}=\frac{a_{1}.(1-q^{n})}{(1-q)}$

When $n$ approaches $\infty$ , if the progression ratio is smaller than 1, the summation has a convergence value. Its value is:

$\zeta=\frac{a_{1}}{(1-q)}$

From the statement, we have:

$\zeta=\frac{2.tg(\frac{\pi}{5})}{1-tg^{2}(\frac{\pi}{5})}$

That's an trigonometric identity, which gives us the tangent of the double angle. Hence, $\zeta=tg (\frac{2\pi}{5})$ , and $s+r=2+5=\boxed{7}$

Trig sum

The answer is 7.

1 solution

0 pending reports