If the $(r+1)^{th}$ term contains the same power of $a$ and $b$ in the expansion of $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ find the value of $r$
I simply applied binomial and expanded for the general term and then equated the power of $a$ and $b$ and got $r=12$
My friend got $r=9$
I want to know which answer is correct.
Please help!!
On
On simplifying the following expression: $$\bigg(\sqrt[3]{\frac{a}{\sqrt b}}+\sqrt{\frac {b}{\sqrt[3]{a}}}\bigg)^{21}$$ We get the expression: $$\frac{1}{(ab)^\frac72}\bigg(a^\frac12+b^\frac23\bigg)^{21}$$
So the (r$+1$)th term of the expansion is $$\binom{n}{r}a^{\frac{21-r-7}{2}}b^{\frac{4r-21}{6}}$$
Hence, the answer will come as: $$\frac{21-r-7}{2}=\frac{4r-21}{6}$$ $$\Rightarrow 42-3r=4r-21$$ $$\Rightarrow \boxed{\color{red}{r=9}}$$
The required answer is $9$.
Without trying to replicate your calculations, you're probably using version of the binomial theorem that count the terms from the opposite end. You can state the theorem either as $$ (p+q)^n = \sum_{r=0}^n \binom{n}{r} p^r q^{n-r} $$ or as $$ (p+q)^n = \sum_{r=0}^n \binom{n}{r} p^{n-r} q^r $$ which both give the same terms, just with a different numbering.
Note in particular that $9+12=21$, so you can both be right, if you're counting from different ends of the expansion!