Binomial expansion by factorial
While we can use either Pascal’s triangle or the
This is useful if we have unknowns in the expression for
Finding a specific term using factorials
If we want to find
Prove that ^5C_2 = 10 using factorials.
^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \times 3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10
Finding multiple coefficients
Expand (1+x)^4 using factorials to find the coefficients
(1+x)^4 = {^4C_0 (1)^4 (x)^0} + {^4C_1 (1)^3 (x)^1} + {^4C_2 (1)^2 (x)^2} + {^4C_3 (1)^1 (x)^3} + {^4C_4 (1)^0 (x)^4} = \frac{4!}{0!(4-0)!} \times 1 + \frac{4!}{1!(4-1)!} \times x + \frac{4!}{2!(4-2)!} \times x^2 + \frac{4!}{3!(4-3)!} \times x^3 + \frac{4!}{4!(4-4)!} \times x^4 = 1 + 4x + 6x^2 + 4x^3 + 1x^4 = 1 + 4x + 6x^2 + 4x^3 + x^4
Find the first 3 terms of the expansion of (2+x)^5 using factorials to find the coefficients
(2+x)^5 = {^5C_0 (2)^5 (x)^0} + {^5C_1 (2)^4 (x)^1} + {^5C_2 (2)^3 (x)^2} + ... = \frac{5!}{0!(5-0)!} \times (2)^5 + \frac{5!}{1!(5-1)!} \times (2)^4 \times x + \frac{5!}{2!(5-2)!} \times (2)^3 \times x^2 + ... = 1 \times 32 + 5 \times 16 \times x + 10 \times 8 \times x^2 + ... = 32 + 80x + 80x^2 + ...
flashcards
| Question | Answer |
|---|---|
| What is the factorial formula for finding | |
| Prove that | |
| Expand | |
| How do you find the first 3 terms of | |
| What is the term for |