Question

a) Determine the middle term in the expansion of \(\left(a-3 b^{3}\right)^{8}\). b) Determine the term containing \(x^{11}\) in the expansion of \(\left(x^{2}-\frac{1}{x}\right)^{10}\).

Short answer

a) The middle term is \(5670 a^{4} b^{12} \). b) The term containing \( x^{11} \) is \( -120 x^{11} \).

Step-by-step solution

Identify the middle term in the binomial expansion

The general term in the expansion of \( (a - 3b^3)^8 \) is given by \( T_{r+1} = \binom{8}{r} a^{8-r} (-3b^3)^r \). Since the power is 8, there are 9 terms. The middle term is the 5th term (as \(\frac{9 + 1}{2} = 5\)).

Substitute values for the middle term

For r = 4 (the 5th term), substitute into the general term formula: \( T_{5} = \binom{8}{4} a^{8-4} (-3b^3)^4 \).

Simplify the middle term

Calculate \( T_{5} \): \( T_{5} = \binom{8}{4} a^4 (-3b^3)^4 = \binom{8}{4} a^4 (81b^{12}) = 70 a^4 81 b^{12} = 5670 a^4 b^{12} \). So, the middle term is \( 5670 a^4 b^{12} \).

Identify the term containing x^11

The general term in the expansion of \( (x^2 - \frac{1}{x})^{10} \) is given by \(T_{r+1} = \binom{10}{r} (x^2)^{10-r} \left(-\frac{1}{x}\right)^r \).

Find the exponent of x

We need the term where the exponent of x is 11. Simplifying the general term's exponent of x: \( x^{2(10-r)} \left(-\frac{1}{x}\right)^r = x^{20-2r} x^{-r} = x^{20-3r} \).

Determine r for x^11

Set \( 20 - 3r = 11 \) and solve for r: \( 3r = 9 \), so \( r = 3 \).

Substitute r to find the specific term

Substitute \( r = 3 \) into the general term formula: \( T_{4} = \binom{10}{3} (x^2)^{10-3} \left(-\frac{1}{x}\right)^3 \).

Simplify the term for x^11

Calculate \( T_{4} \): \(\binom{10}{3} (x^2)^7 (-\frac{1}{x})^3 = \binom{10}{3} x^{14} \left(-\frac{1}{x}\right)^3 = \binom{10}{3} x^{14} (-\frac{1}{x^3}) = \binom{10}{3} (-1) x^{11} = 120 (-1) x^{11} = -120 x^{11} \). Thus, the term containing \(x^{11} \) is \( -120 x^{11} \).

Key concepts

binomial theorem

The binomial theorem is a fundamental formula in algebra. It helps to expand expressions that are raised to a power, like \( (a+b)^n \). The theorem states that this expression can be expanded into a series of terms involving coefficients, powers of 'a,' and powers of 'b.' The general form is:
\[ (a+b)^n = \binom{n}{0} a^n + \binom{n}{1} a^{n-1} b + \binom{n}{2} a^{n-2} b^2 + ... + \binom{n}{n} b^n \]
Each term has a binomial coefficient \[ \binom{n}{r} \] which counts the number of ways to choose 'r' elements from a set of 'n' elements.

general term formula

To find a specific term in the expansion without writing all terms, we use the general term formula. The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here’s how it works:

• \( \binom{n}{r} \) is the binomial coefficient.
• 'a' is raised to the power of \( n-r \).
• 'b' is raised to the power of 'r'.

This formula lets us directly calculate any term in the expansion.

exponents in binomial expansions

The exponents in binomial expansions follow predictable patterns. In the general term \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \), we see how the exponents of 'a' and 'b' change.

• The exponent of 'a' decreases from 'n' to '0'.
• The exponent of 'b' increases from '0' to 'n' as 'r' increases from 0 to n.

For example, the exponents of 'x' in \( (x^2 - \frac{1}{x})^{10} \) require careful handling. Here, each term's x-elements’ exponents must be combined to get the resulting term.

combination formula

The combination formula, denoted as \( \binom{n}{r} \), calculates how many ways we can choose 'r' elements from 'n' elements without regard to the order. It’s essential in binomial expansion. The formula is:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Where \( n! \) (n factorial) is the product of all positive integers up to 'n'. For example, for \( n = 8 \) and \( r = 4 \), \[ \binom{8}{4} = \frac{8!}{4!(8-4)!} = 70 \]
Combination values determine the coefficients in the binomial expansion, which are crucial for correctly forming each term.