Question

\(\left[\left(a^{1 / 3} / b^{1 / 6}\right)+\left(b^{1 / 2} / a^{1 / 6}\right)\right]^{21}\) 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

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

Step-by-step solution

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").

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: \(\binom{21}{r} \left(a^{1 / 3}\right)^{21-r} \left(b^{1 / 6}\right)^r \left(b^{1 / 2}\right)^r \left(a^{1 / 6}\right)^r\).

Simplify the general term

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

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 \(b\): \(7 - \frac{r}{6} = \frac{2r}{3}\)

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.

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.

Correct the general term

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

Simplify the corrected general term

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

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

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

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.

Key concepts

Exponentiation

Exponentiation involves raising numbers to a given power. It is crucial in algebra, especially when simplifying expressions or solving equations. In our exercise, we see exponents like \(a^{1/3}\), \(b^{1/6}\), and \(21\), forming expressions elevated to powers. Exponentiation tells us how many times a number, tied to a base, multiplies itself.

• For example, \(a^{1/3}\) denotes the cube root of \(a\).
• The key principle is: the larger the exponent, the higher the multiplication power applied to the base.

Using exponentiation rules, such as multiplication \((x^m \cdot x^n = x^{m+n})\) or power to a power \((x^{m})^{n} = x^{m\cdot n}\), we can transform expressions. This helps identify relationships between variables like \(a\) and \(b\) in the binomial terms of the equation.

Polynomial Expansion

Polynomial expansion involves expressing a given polynomial to demonstrate its individual terms. This often uses the binomial theorem, which provides a formulaic way to expand polynomials raised to any power.
In our problem, the expression \[\left[ \left( a^{1/3} / b^{1/6} \right) + \left( b^{1/2} / a^{1/6} \right) \right]^{21}\]is expanded using the binomial theorem. This principle allows us to break down complex expressions into sums of simpler, individual terms.

• Each term in the expansion represents a coefficient, expression, and degree.
• This approach provides a systematic way of distributing powers and coefficients across each term.

When tackling polynomial expansions, we focus on finding specific terms like the \((r+1)\)th term in this exercise, achieved by applying the binomial coefficient formula tightly linked to exponentiation rules.

Algebraic Expressions

These are combinations of numbers, variables, and operators (like +, -, *, /) that come together to express a mathematical statement. Understanding how these expressions work is vital for solving equations. In our case, expressions involve fractions, roots, and powers of \(a\) and \(b\).

• We manipulate these expressions using algebraic rules to simplify or rearrange them.
• It involves balancing equations, combining like terms, and transforming them as needed.

For example, simplifying expressions like \(a^{7 - r/6}\) and \(b^{2r/3}\) by setting their powers equal helps uncover meaningful relationships between the variables.
This knowledge enables us to approach binomial problems analytically, efficiently navigating through complex exponential terms.