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Chapter 7: Problem 568

\(\left[\left(a^{1 / 3} / b^{1 / 6}\right)+\left(b^{1 / 2} / a^{1 / 6}\right)\right]^{21}\) has same power of a and \(b\) for \((\mathrm{r}+1)\) th term then \(\mathrm{r}=\) (a) 8 (b) 9 (c) 10 (d) 11

Short Answer

Expert verified
None of the provided options are correct, as the correct value of \(r\) is 0.

Step by step solution

01

Write down the binomial theorem

The binomial theorem states that: \((x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\), where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) (pronounced as "n choose k").
02

Find the general (r+1)th binomial term

Using the binomial theorem, the general binomial term, i.e., the \((r+1)\) th term of the given expression, when \(n = 21\) is given by: \(\binom{21}{r} \left(a^{1 / 3}\right)^{21-r} \left(b^{1 / 6}\right)^r \left(b^{1 / 2}\right)^r \left(a^{1 / 6}\right)^r\).
03

Simplify the general term

Now, let's simplify the powers of \(a\) and \(b\) in the general term: \(a^{(21-r)(1/3) + r(1/6)} \cdot b^{r(1/6) + r(1/2)} = a^{7 - \frac{r}{3} + \frac{r}{6}} \cdot b^{\frac{r}{6} + \frac{r}{2}} = a^{7-\frac{r}{6}} \cdot b^{\frac{2r}{3}}\)
04

Equate the powers of a and b

We need to find \(r\) such that the powers of \(a\) and \(b\) are equal in the \((r+1)\)th term. So, we equate the powers of \(a\) and \(b\): \(7 - \frac{r}{6} = \frac{2r}{3}\)
05

Solve for r

Now, we solve the equation for \(r\): \(7 - \frac{r}{6} = \frac{2r}{3}\) Multiplying both sides by 6 to eliminate fractions: \(42 - r = 4r\) Adding \(r\) to both sides: \(42 = 5r\) Dividing by 5: \(r = \frac{42}{5}\) However, \(r\) must be an integer, and this solution does not satisfy this condition.
06

Re-examine the problem

Based on our analysis, we might have made a mistake in writing the general term. Re-checking our work, we see that in step 2, we should have multiplied the powers of \(a\) and \(b\) in the general term, not added them.
07

Correct the general term

The corrected general term should be: \(\binom{21}{r} \left(a^{1 / 3}\right)^{21-r} \left(b^{1 / 6}\right)^r \left(b^{1 / 2}\right)^{21-r} \left(a^{1 / 6}\right)^{21-r}\).
08

Simplify the corrected general term

Simplifying the corrected general term: \(a^{(21-r)(1/3) + (21-r)(1/6)} \cdot b^{r(1/6) + (21-r)(1/2)} = a^{7 + \frac{r}{6}} \cdot b^{\frac{2r+21}{3}}\)
09

Equate the powers of a and b in the corrected general term

Equating the powers of \(a\) and \(b\) in the corrected general term: \(7+\frac{r}{6} = \frac{2r+21}{3}\)
10

Solve the equation for r

Multiplying both sides by 6 to eliminate fractions: \(42+6r = 4r+42\) Subtracting \(4r\) and \(42\) from both sides: \(6r - 4r = 42 - 42\) Simplifying: \(2r=0\) Dividing by 2: \(r=0\) So, the \((r+1)\) th term corresponds to the first term in the expansion. Answer: None of the provided options are correct, as the correct value of \(r\) is 0.

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