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# Find the number of rational terms in the expansion of ${{({{9}^{\dfrac{1}{4}}}+{{8}^{\dfrac{1}{6}}})}^{1000}}$ .

Last updated date: 13th Jun 2024
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Hint: General binomial expansion of ${{(a+b)}^{n}}={}^{n}{{c}_{r}}{{(a)}^{n-r}}{{(b)}^{r}}$ where
${}^{n}{{c}_{r}}=\dfrac{n!}{(n)!(n-r)!}$
Here ${}^{n}{{c}_{r}}$ term is always a rational number, so we want the terms ${{(a)}^{n- r}}and{{(b)}^{r}}$ to be rational.

For making these terms rational, we have to make $n-r$ and $r$ as an integer, because non
integer power to an integer number can never be an integer.
So basically, we want to make the powers of a and b as integers.
We are given a binomial expression as ${{({{9}^{\dfrac{1}{4}}}+{{8}^{ \dfrac{1}{6}}})}^{1000}}$
Using formula ${{(a+b)}^{n}}={}^{n}{{c}_{r}}{{(a)}^{n-r}}{{(b)}^{r}}$, here r varies from 0 to n
Expanding,
${{({{9}^{\dfrac{1}{4}}}+{{8}^{\dfrac{1}{6}}})}^{1000}}={}^{1000}{{c}_{r}}{{({{9}^{\dfrac{1}{4}}})}^{1000 -r}}{{({{8}^{\dfrac{1}{6}}})}^{r}}$, similarly r varies from 0 to 1000
now as we know that $9={{3}^{2}}$ and $8={{2}^{3}}$ so we can replace their value in above
equation 
we can write ${{9}^{\dfrac{1}{4}}}={{({{3}^{2}})}^{\dfrac{1}{4}}}$ and ${{8}^{ \dfrac{1}{6}}}={{({{2}^{3}})}^{\dfrac{1}{6}}}$
Using property ${{({{x}^{a}})}^{b}}={{x}^{ab}}$
We can write ${{9}^{\dfrac{1}{4}}}=({{3}^{\dfrac{2}{4}}})={{3}^{\dfrac{1}{2}}}...(2)$ and ${{8}^{ \dfrac{1}{6}}}=({{2}^{\dfrac{3}{6}}})={{2}^{\dfrac{1}{2}}}.....(3)$
Substituting equation (2) and (3) in equation (1)
${{({{9}^{\dfrac{1}{4}}}+{{8}^{ \dfrac{1}{6}}})}^{1000}}={}^{1000}{{c}_{r}}{{({{3}^{\dfrac{1}{2}}})}^{1000- r}}{{({{2}^{\dfrac{1}{2}}})}^{r}}$
Again, using property ${{({{x}^{a}})}^{b}}={{x}^{ab}}$
We can write it as ${{({{9}^{\dfrac{1}{4}}}+{{8}^{ \dfrac{1}{6}}})}^{1000}}={}^{1000}{{c}_{r}}({{3}^{\dfrac{1000-r}{2}}})({{2}^{\dfrac{r}{2}}})$
\begin{align} \to {}^{1000}{{c}_{r}}({{3}^{500-\dfrac{r}{2}}})({{2}^{\dfrac{r}{2}}}) \\ \\ \end{align}
Now as we know that ${}^{1000}{{c}_{r}}$ is integer so we just want powers of 2 and 3 to be
integer to give rational term, and looking carefully we just want $\dfrac{r}{2}$ to be integer ,
because $500-\dfrac{r}{2}$ will also become integer if $\dfrac{r}{2}$ is integer.
Now what if we take $\dfrac{r}{2}$ as non integer for ex r=3 , then we can see that it becomes
${{2}^{\dfrac{3}{2}}}$ so it’s clearly not an rational number
We just want $\dfrac{r}{2}$ to be integer and our r varies from 0 to 1000
So $\dfrac{r}{2}$ will be integer whenever r will be multiple of 2
Values of r = 0,2,4….1000
Which is equals to $\dfrac{1000}{2}+1=501$
Hence 501 terms are rational.

Note: You can do some mistake while expanding the binomial expression or In using the property
${{({{x}^{a}})}^{b}}={{x}^{ab}}$ correctly, Convince yourself that power of a integer number must be
an integer to give a rational number , if you have a doubt cross check it by putting any non-
integer number in power of any integer number for ex-
${{2}^{\dfrac{1}{3}}}$ or ${{3}^{\dfrac{3}{8}}}$ they can’t be a rational number.