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

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Last updated date: 13th Jun 2024
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Answer
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411.9k+ views
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.

Complete step-by-step answer:
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.