Question

Show that the sum of the first $n$ natural numbers is a perfect square, if $n$ is equal to $k^2$ or ${k^{{}^{\prime }}}^2-1$, where $k$ is the numerator of an odd and $k^{{}^{\prime }}$ the numerator of an even convergent to $\sqrt{2}$.

Answer 1

The sum of lot '$n$' natural number is $\frac{n(n+1)}{2}$

This expression is a perfect square when $n=k^2$, provided that $\frac{k^2+1}{2}$ is a perfect square$=x^2$(say)

$\Rightarrow k^2-2x^2+1=0$

$\Rightarrow \sqrt{2}=1+\frac{1}{2+}\frac{1}{2+}\frac{1}{2+}.....$

The values of $k$ are the numerators of the odd convergents, since the no. of quotients in the period is odd.

Again, the expression $\frac{n(n+1)}{2}$ is a perfect square when $n+1=k^{{{}^{\prime }}^2}$ provided that $\frac{k^{{{}^{\prime }}^2}-1}{2}=a$a perfect square$=x^2$.

$\Rightarrow k^{{{}^{\prime }}^2}-2x^2=1$

In this case the roots of equations are the numerators of the even convergents.