0
$\begingroup$

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!!

$\endgroup$
2
$\begingroup$

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!

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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 $9$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ couldn't it be 12 as 12+9=21? $\endgroup$ – Atul Mishra Jul 8 '17 at 10:56
  • $\begingroup$ The first step here looks somewhat obscure. Why not simplify to $$ (a^{1/3}b^{-1/6}+a^{-1/6}b^{1/2})^{21}$$ and then solve $$ \frac13r - \frac16(21-r) = -\frac16 r + \frac12(21-r) $$ $\endgroup$ – hmakholm left over Monica Jul 8 '17 at 10:59
  • $\begingroup$ @AtulMishra Well, in a problem, where the binomial expression is rigorously given, it is expected that you consider the first term as the 'a' term and the second one 'b'. So if you consider the expansion in this order, your answer should be 9. I saw Henning's answer, but I dont think that you are allowed to reverse the expression and work. It may seem trivial that you cannot reverse the expression but if one is allowed to do so, then the ordering of the terms will become ambiguous and there will be no unique answer. Hope this helps. $\endgroup$ – SchrodingersCat Jul 8 '17 at 11:02
  • $\begingroup$ Yes I am just arguing for the same, It will give r=12 $\endgroup$ – Atul Mishra Jul 8 '17 at 11:03
  • $\begingroup$ @SchrodingersCat: There's no "reversing" going on -- there's just two ways to state the binomial theorem which are exactly equally good. It's not as if one of them is "the right way" and the other is "reversed". $\endgroup$ – hmakholm left over Monica Jul 8 '17 at 11:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.