Question

Find m if the following equation holds true $2+4+6+...+2n=m^2\times n(n+1)$

Answer 1

$2+4+6+................+2n=m^2\times n(n+1)$

$2(1+2+3+..............+n)=m^2\times n(n+1)$

$2\times \frac{n(n+1)}{2}=m^2\times n(n+1)$ ( Sum of first $n$ natural numbers $=\frac{n(n+1)}{2}$ )

$n(n+1)=m^2\times n(n+1)$

$\therefore m^2=1$

$\therefore m=\pm 1$