| 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.
Correct Answer: A
Solution :
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\[-2.\]
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