Question

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

Answer 1

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,

-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{aligned} \to {}^{1000}{{c}_{r}}({{3}^{500-\dfrac{r}{2}}})({{2}^{\dfrac{r}{2}}}) \\\ \\\ \end{aligned}$$ Now as we know that $${}^{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.