Question

pascals triangle & binomial theorem ws #2
name
group

1. expand $(2x-3y)^3$
2. expand $(x+4y)^4$
3. given $(a+b)^7$, find the term containing $a^4b^3$
4. given $(a+b)^7$, find the sixth term.

Explanation:

Problem 1: Expand $(2x-3y)^3$

Step1: Recall binomial formula

For $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, let $a=2x$, $b=3y$.

Step2: Substitute $a$ and $b$

$(2x)^3 - 3(2x)^2(3y) + 3(2x)(3y)^2 - (3y)^3$

Step3: Compute each term

$8x^3 - 3(4x^2)(3y) + 3(2x)(9y^2) - 27y^3$

Step4: Simplify the expression

$8x^3 - 36x^2y + 54xy^2 - 27y^3$

---

Problem 2: Expand $(x+4y)^4$

Step1: Recall binomial formula

For $(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$, let $a=x$, $b=4y$.

Step2: Substitute $a$ and $b$

$x^4 + 4x^3(4y) + 6x^2(4y)^2 + 4x(4y)^3 + (4y)^4$

Step3: Compute each term

$x^4 + 16x^3y + 6x^2(16y^2) + 4x(64y^3) + 256y^4$

Step4: Simplify the expression

$x^4 + 16x^3y + 96x^2y^2 + 256xy^3 + 256y^4$

---

Problem 3: Find $a^4b^3$ term in $(a+b)^7$

Step1: Recall general binomial term

The term with $a^{n-k}b^k$ is $\binom{n}{k}a^{n-k}b^k$, $n=7$, $k=3$.

Step2: Calculate binomial coefficient

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = 35$

Step3: Write the final term

$35a^4b^3$

---

Problem 4: Find 6th term in $(a+b)^7$

Step1: Identify term index

For the $r$-th term, $k=r-1$. Here $r=6$, so $k=5$, $n=7$.

Step2: Calculate binomial coefficient

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = 21$

Step3: Write the final term

$21a^{7-5}b^5 = 21a^2b^5$

Answer:

1. $8x^3 - 36x^2y + 54xy^2 - 27y^3$
2. $x^4 + 16x^3y + 96x^2y^2 + 256xy^3 + 256y^4$
3. $35a^4b^3$
4. $21a^2b^5$