Show that the integral part of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ is odd if $n$ be a positive integer.

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Hint: For solving this question first we will expand the given term using a binomial expansion formula then we will try to figure out whether the integral part (i.e part that involves no fractions) is odd or not.

Complete step by step answer:
We have to prove that the integral part of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ is odd if $n>0$ .
Now, we will use the following binomial expansion result:
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}\cdot y+{}^{n}{{C}_{2}}{{x}^{n-2}}\cdot {{y}^{2}}+{}^{n}{{C}_{3}}{{x}^{n-3}}\cdot {{y}^{3}}+{}^{n}{{C}_{4}}{{x}^{n-4}}\cdot {{y}^{4}}+{}^{n}{{C}_{5}}{{x}^{n-5}}\cdot {{y}^{5}}+....................+{}^{n}{{C}_{n}}{{y}^{n}}\]
Where, ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
We can apply the above result to expand ${{\left( 5+2\sqrt{6} \right)}^{n}}$ . Then,
${{\left( 5+2\sqrt{6} \right)}^{n}}={}^{n}{{C}_{0}}{{5}^{n}}+{}^{n}{{C}_{1}}{{5}^{n-1}}\left( 2\sqrt{6} \right)+{}^{n}{{C}_{2}}{{5}^{n-2}}{{\left( 2\sqrt{6} \right)}^{2}}+{}^{n}{{C}_{3}}{{5}^{n-3}}{{\left( 2\sqrt{6} \right)}^{3}}+{}^{n}{{C}_{4}}{{5}^{n-4}}{{\left( 2\sqrt{6} \right)}^{4}}+........+{}^{n}{{C}_{n}}{{\left( 2\sqrt{6} \right)}^{n}}$
We can also write it like, ${{\left( 5+2\sqrt{6} \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{5}^{n-r}}{{\left( 2\sqrt{6} \right)}^{r}}}$ .
Now, if you see the expansion of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ then, you can see that for the integral part value of $r$ should be even like $r=0,2,4,6,8$ till the maximum even number less than or equal to $n$ . We can write that, ${{\left( 5+2\sqrt{6} \right)}^{n}}=I+F$ where $I$ is the integral part and $F$ is the irrational part. Then,
  & I={}^{n}{{C}_{0}}{{5}^{n}}+{}^{n}{{C}_{2}}{{5}^{n-2}}{{\left( 2\sqrt{6} \right)}^{2}}+{}^{n}{{C}_{4}}{{5}^{n-4}}{{\left( 2\sqrt{6} \right)}^{4}}+{}^{n}{{C}_{6}}{{5}^{n-6}}{{\left( 2\sqrt{6} \right)}^{6}}+....................... \\
 & \Rightarrow I={{5}^{n}}+{}^{n}{{C}_{2}}{{5}^{n-2}}\left( {{2}^{2}}\times 6 \right)+{}^{n}{{C}_{4}}{{5}^{n-4}}\left( {{2}^{4}}\times {{6}^{2}} \right)+{}^{n}{{C}_{6}}{{5}^{n-6}}\left( {{2}^{6}}\times {{6}^{3}} \right)+...................... \\
 & \Rightarrow I={{5}^{n}}+2\left[ {}^{n}{{C}_{2}}{{5}^{n-2}}\left( {{2}^{1}}\times 6 \right)+{}^{n}{{C}_{4}}{{5}^{n-4}}\left( {{2}^{3}}\times {{6}^{2}} \right)+{}^{n}{{C}_{6}}{{5}^{n-6}}\left( {{2}^{5}}\times {{6}^{3}} \right)+......................... \right] \\
 & \Rightarrow I={{5}^{n}}+2k \\
Where, $k={}^{n}{{C}_{2}}{{5}^{n-2}}\left( {{2}^{1}}\times 6 \right)+{}^{n}{{C}_{4}}{{5}^{n-4}}\left( {{2}^{3}}\times {{6}^{2}} \right)+{}^{n}{{C}_{6}}{{5}^{n-6}}\left( {{2}^{5}}\times {{6}^{3}} \right)+.........................$ . And the value of $2k$ will always be even. Moreover, the value of ${{5}^{n}}$ will be always odd for any positive value of $n$ .
Now, we can write that integral part of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ is the sum of one odd number and even number. And as we know that sum of one odd number and one even number will be odd always. So, the integral part of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ will be odd always for any positive value of $n$ .
Hence, proved.

Note: Here, the student should write the expansion of ${{\left( 5+2\sqrt{6} \right)}^{n}}$ correctly as per the binomial expansion formula without missing any term. Then, check each term clearly whether it is rational or irrational then analyse the integral part to prove the desired result without any calculation mistake.