Mathematics - Binomial Theorem Question with Solution | TestHub

We know the binomial expansion of ${\left(a+b\right)}^n$ is,

${\left(a+b\right)}^n={}^{n}C_0{a}^n{b}^0+{}^{n}C_1{a}^{n-1}{b}^1+{}^{n}C_2{a}^{n-2}{b}^2+......+{}^{n}C_n{a}^0{b}^n$

Using the above expansion,

$x={\left(8\sqrt{3}+13\right)}^{13}={}^{13}C_0{\left(8\sqrt{3}\right)}^{13}+{}^{13}C_1{\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^1+\dots$

Let $x^{\text{'}}={\left(8\sqrt{3}-13\right)}^{13}={}^{13}\mathrm{C}_0{\left(8\sqrt{3}\right)}^{13}-{}^{13}\mathrm{C}_1{\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^1+\dots$

So, $x-x^{\text{'}}=2\left[{}^{13}C_1·{\left(8\sqrt{3}\right)}^{12}{\left(13\right)}^1+{}^{13}C_3{\left(8\sqrt{3}\right)}^{10}·{\left(13\right)}^3\dots \right]$

Therefore, $x-x^{\text{'}}$ is an even integer, hence $\left[x\right]$ is even

Now, $y={\left(7\sqrt{2}+9\right)}^9={}^9\mathrm{C}_0{\left(7\sqrt{2}\right)}^9+{}^9\mathrm{C}_1{\left(7\sqrt{2}\right)}^8{\left(9\right)}^1+{}^9\mathrm{C}_2{\left(7\sqrt{2}\right)}^7{\left(9\right)}^2+....$Now let,${\mathrm{y}}^{\text{'}}=(7\sqrt{2}-9{)}^9={}^9\mathrm{C}_0(7\sqrt{2}{)}^9-{}^9\mathrm{C}_1\left(7\sqrt{2}{)}^8\right(9{)}^1+{}^9\mathrm{C}_2{\left(7\sqrt{2}\right)}^7{\left(9\right)}^2$

So, $y-y^{\text{'}}=2\left[{}^{9}C_1{\left(7\sqrt{2}\right)}^8{\left(9\right)}^1+{}^{9}C_3{\left(7\sqrt{2}\right)}^6{\left(9\right)}^3+\dots \right]$

$y-y^{\text{'}}=$ Even integer hence $\left[\mathrm{y}\right]$ is even