• # question_answer Statement-1: If $n$ is an odd prime, then integral part of${{(\sqrt{5}+2)}^{n}}$is divisible by$n$. Statement-2: If $n$ is prime, then$^{n}{{c}_{1}},\,{{\,}^{n}}{{c}_{2}},\,{{\,}^{n}}{{c}_{3}},...{{,}^{n}}{{c}_{n-1}}$, must be divisible by$n$. A)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.B)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.C)  Statement-1 is true, Statement-2 is false.D)  Statement-1 is false, Statement-2 is true.

Let${{\left( \sqrt{5}+2 \right)}^{n}}=I+f$, where$I$is an integer and$0<f<1$ Let,${{\left( \sqrt{5}-2 \right)}^{n}}=f';\,\,0<f'<1$ $\Rightarrow$${{\left( \sqrt{5}+2 \right)}^{n}}-{{\left( \sqrt{5}-2 \right)}^{n}}=$Integer$(\because \,\,n$is odd) $\because$${{\left( \sqrt{5}+2 \right)}^{n}}-{{\left( \sqrt{5}-2 \right)}^{n}}=2\left[ ^{n}{{c}_{1}}{{2.5}^{\frac{n-1}{2}}}{{+}^{n}}{{c}_{3}}{{2}^{3}}{{.5}^{\frac{n-3}{2}}}+... \right]$$\because$$(f=f')$ $\Rightarrow$$I$is divisible by $20n$ on using statement$-2.$