Question

c10 binomial theorem & pascals triangle ws#1 name group 3

1. expand $(x+y)^4$
2. expand $(3x-2y)^3$
3. given $(a+b)^{12}$, find the term containing $a^5b^7$
4. given $(a+b)^{12}$, find the fourth term.

Explanation:

Step1: Apply Binomial Theorem to $(x+y)^4$

The binomial theorem states $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$. For $n=4$:
$$(x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3 + \binom{4}{4}x^0y^4$$
Calculate coefficients: $\binom{4}{0}=1$, $\binom{4}{1}=4$, $\binom{4}{2}=6$, $\binom{4}{3}=4$, $\binom{4}{4}=1$
Simplify: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Step2: Apply Binomial Theorem to $(3x-2y)^3$

For $n=3$, $a=3x$, $b=-2y$:
$$(3x-2y)^3 = \binom{3}{0}(3x)^3(-2y)^0 + \binom{3}{1}(3x)^2(-2y)^1 + \binom{3}{2}(3x)^1(-2y)^2 + \binom{3}{3}(3x)^0(-2y)^3$$
Calculate coefficients and terms:
$\binom{3}{0}(3x)^3 = 1\cdot27x^3=27x^3$
$\binom{3}{1}(3x)^2(-2y) = 3\cdot9x^2\cdot(-2y)=-54x^2y$
$\binom{3}{2}(3x)(-2y)^2 = 3\cdot3x\cdot4y^2=36xy^2$
$\binom{3}{3}(-2y)^3 = 1\cdot(-8y^3)=-8y^3$
Combine: $27x^3 - 54x^2y + 36xy^2 - 8y^3$

Step3: Find term with $a^5b^7$ in $(a+b)^{12}$

Use formula $\binom{n}{k}a^{n-k}b^k$, here $n-k=5$, $k=7$, $n=12$
Calculate coefficient: $\binom{12}{7} = \frac{12!}{7!(12-7)!} = \frac{12\times11\times10\times9\times8}{5\times4\times3\times2\times1}=792$
Term: $792a^5b^7$

Step4: Find 4th term of $(a+b)^{12}$

The $r$-th term is $\binom{n}{r-1}a^{n-(r-1)}b^{r-1}$, $r=4$, $n=12$
$r-1=3$, so coefficient $\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12\times11\times10}{3\times2\times1}=220$
Term: $\binom{12}{3}a^{12-3}b^3 = 220a^9b^3$

Answer:

1. $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$
2. $27x^3 - 54x^2y + 36xy^2 - 8y^3$
3. $792a^5b^7$
4. $220a^9b^3$