If $a$ be the amplitude of each then $y = a\sin 2\pi (n - 1)t,$ $y = a\sin 2\pi nt$ and ${y_3} = a\sin 2\pi (n + 1)t$
Resultant displacement due to all three waves is $y = {y_1} + {y_2} + {y_3}$
$ = a\sin 2\pi nt + a\,[\sin 2\pi (n - 1)t + \sin 2\pi (n + 1)t]$
$ = a\sin 2\pi nt + a\,[2\sin 2\pi nt\cos 2\pi t]$
$\left[ {{\rm{Using }}\sin C + \sin D = 2\sin \frac{{C + C}}{2}\cos \frac{{C - D}}{2}} \right]$
==> $y = a\,(1 + \cos 2\pi t)\sin 2\pi nt$
This is the resultant wave having amplitude $ = (1 + \cos 2\pi t)$
For maximum amplitude $\cos 2\pi t = 1 ==> 2\pi t = 2m\pi \,\,$ where $m = 0, 1, 2, 3, ...$
$==> t = 0, 1, 2, 3 ...$
Hence time interval between two successive maximum is $1 sec.$ So beat frequency $= 1$
Also for minimum amplitude $(2\cos 2\pi  t) = 0$
==> $\cos 2\pi t = - \frac{1}{2}$
==> $2\pi t = 2m\pi + \frac{{2\pi }}{3}$ ==> $t = + \frac{1}{3}$
==> $t = \frac{1}{3},\frac{4}{3},\frac{7}{3},\frac{{10}}{3},....$ (for $m = 0, 1, 2, ..$)
Hence time interval between two successive minima is $1 sec$ so, number of beats per second $= 1$