Question

Assertion : If n is an odd prime then $\left[{(\sqrt{5}+2)}^n\right]-2^{n+1}$ is not divisible by $20n$, where [.] denotes greatest integer function.

Reason: If n is prime then ${}^n{C}_1{,}^n{C}_2....{.}^n{C}_{n-1}$ must be divisible by $n$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is D Both Assertion and Reason are incorrect

We have ${}^P{C}_r=\frac{P!}{r!(p-r)!}\Rightarrow r!(P-r){!}^P{C}_r=P!$
As $P|P!$ we get $P|r!{\left(P\_r\right)}^P{C}_r$
But for $1\le r\le P-1$, neither $r!$ nor $(P-r)!$is divisible by $P$
$\therefore P{|}^P{C}_r$
We have $\sqrt{5}-2=\frac{1}{\sqrt{5}+2}\Rightarrow 0<\sqrt{5}-2<1\Rightarrow 0<f={(\sqrt{5}-2)}^n<1$
Let ${(2+\sqrt{5})}^n+N+F$ where $0<F<1$
Now $N+F-f-2^{n+1}={(\sqrt{5}+2)}^n-{(\sqrt{5}-2)}^n$
$=2\left[{}^n{C}_1{s}^{\frac{(n-1)}{2}}\left(2\right){+}^n{C}_3{s}^{\frac{n-3}{2}}{2}^3\right]+...{+}^n{C}_{n-2}\left(5\right)2^{n-1}$ (1)
Since $n$ is odd
Since $n$ is an odd prime each of ${}^n{C}_1{,}^n{C}_3,..{.}^n{C}_{n-2}$ is divisible by n
Thus R.H.S of (1) is divisible by $20n$
Also $F-f$ is an integer
Since $0<F<1$ and $0<f<1$, we get $-1<F-f<1$
As $F-f$ is an integer, we get $F-f=0$ or $F=f$
Therefore integral part of ${(2+\sqrt{5})}^n-2^{n+1}$ is ${N-2}^{n+1}$ which is divisible by $20n$