Question

*Find the third term in the expansion of (x + 3y)⁵ from the beginning.*

1️⃣ 90x²y³
2️⃣ 180x³y²
3️⃣ 90
4️⃣ 90x³y²​

Answer 1

Answer:

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Answer 2

Answer:

The correct answer is option 4

Step-by-step explanation:

Given: Equation

To find: the third term in the expansion of the equation

Solution:

Use the binomial expansion theorem to find each term. The binomial theorem states $(a+b)^{n}=\sum_{k=0}^{n} n C k \cdot\left(a^{n-k} b^{k}\right)$.

$\sum_{k=0}^{5} \frac{5 !}{(5-k) ! k !} \cdot(x)^{5-k} \cdot(-3 y)^{k}$

Expand the summation.

$\frac{5 !}{(5-0) ! 0 !}(x)^{5-0} \cdot(-3 y)^{0}+\frac{5 !}{(5-1) ! 1 !}(x)^{5-1} \cdot(-3 y)+\ldots+\frac{5 !}{(5-5) ! 5 !}(x)^{5-5} \cdot(-3 y)^{5}$

Simplify the exponents for each term of the expansion.

$1 \cdot(x)^{5} \cdot(-3 y)^{0}+5 \cdot(x)^{4} \cdot(-3 y)+\ldots+1 \cdot(x)^{0} \cdot(-3 y)^{5}$

Simplify the polynomial result.

$x^{5}-15 x^{4} y+90 x^{3} y^{2}-270 x^{2} y^{3}+405 x y^{4}-243 y^{5}$