Question

Statement $1$:
If $p$ is a prime number $(p\ne 2)$, then $\left[\right(2+\sqrt{5}{)}^p]-2^{p+1}$ is always divisible by $p$ (where $[.]$ denotes the greatest integer function).

Statement $2$:
If $n$ is a prime, then ${}^n{C}_1,{}^n{C}_2{,}^n{C}_3,\dots {,}^n{C}_{n-1}$ must be divisible by $n$.

• A. Both the statements are TRUE and STATEMENT $2$ is the correct explanation of STATEMENT $1$.
• B. Both the statements are TRUE and STATEMENT $2$ is NOT the correct explanation of STATEMENT $1$.
• C. STATEMENT $1$ is TRUE and STATEMENT $2$ is FALSE.
• D. STATEMENT $1$ is FALSE and STATEMENT $2$ is TRUE.

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT $2$ is the correct explanation of STATEMENT $1$.

We have,
${(2+\sqrt{5})}^p+{(2-\sqrt{5})}^p=2[2^p+{}^p{C}_2{2}^{p-2}5+{}^p{C}_4{2}^{p-4}{5}^2+\dots +{}^p{C}_{p-1}2\cdot 5^{(p-1)/2}]$
$\dots \left[1\right]$
From $\left[1\right]$,
${(2+\sqrt{5})}^p+{(2-\sqrt{5})}^p$ is an integer and $-1<{(2-\sqrt{5})}^p<0$
($\because p$ is odd)

So, $\left[{(2+\sqrt{5})}^p\right]={(2+\sqrt{5})}^p+{(2-\sqrt{5})}^p$

$=2^{p+1}+{}^p{C}_2{2}^{p-1}5+\dots +{}^p{C}_{p-1}{2}^2\cdot 5^{(p-1)/2}$

$\Rightarrow \left[{(2+\sqrt{5})}^p\right]-2^{p+1}=2[{}^p{C}_2{2}^{p-2}5+{}^p{C}_4{2}^{p-4}{5}^2+\dots +{}^p{C}_{p-1}2\cdot 5^{(p-1)/2}]$
Now, all the binomial coefficients
${}^p{C}_2=\frac{p(p-1)}{1\times 2}$,
${}^p{C}_4=\frac{p(p-1)(p-2)(p-3)}{1\times 2\times 3\times 4},$
$⋮$
${}^p{C}_{p-1}=p$
are divisible by the prime $p$.
Thus, RHS is divisible by $p$.