Question

If ${(7+4\sqrt{3})}^n=s+t$, where $n$ and $s$ are positive integers and $t$ is a proper fraction, then value of $(1-t)(s+t)$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

Since $t$ is a proper fraction $\therefore 0<t<1$ given ${(7+4\sqrt{3})}^n=s+t$

Now, let $t^{\prime }={(7-4\sqrt{3})}^n,0<t^{\prime }<1$

Also $s+t+t^{\prime }={(7+4\sqrt{3})}^n+{(7-4\sqrt{3})}^n=2\{7^n+{{}^{n}C}_2{\left(7\right)}^{n-2}{\left(4\sqrt{3}\right)}^2+{{}^{n}C}_4{\left(7\right)}^{n-4}{\left(4\sqrt{3}\right)}^4+...\}=2\left(Integer\right)=2k(k\in N)=$Even integer

Hence $t+t^{\prime }=$Even integer$-s$, but $0<t+t^{\prime }<2$
$\therefore$ R.H.S is integer, hence L.H.S is also integer. Therefore $t+t^{\prime }=1$

$\therefore t^{\prime }=(1-t)$

Hence $(1-t)(s+t)=t^{\prime }(s+t)=\left[\right(7-4\sqrt{3}\left)\right(7+4\sqrt{3}){]}^n=1$