This question tests your understanding of the Binomial Theorem—a formula for expanding (x + $y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. To find the coefficient of $x^2y$^3 in (x + $y)^5$, we need the term where x has power 2 and y has power 3 (note: 2 + 3 = 5, which matches our exponent). In the expansion, this is the 4th term (counting from term 1), and row 5 of Pascal's Triangle is 1, 5, 10, 10, 5, 1—so the 4th coefficient is 10. Choice B correctly identifies this coefficient as 10, while Choice A (5) would be the coefficient of $x^4y$ or $xy^4$, and Choice C (20) doesn't appear in row 5 at all. Quick Pascal's Triangle construction: write 1s down both edges, then for interior numbers, add the two directly above—row 5 builds from row 4 (1, 4, 6, 4, 1) to get 1, 1+4=5, 4+6=10, 6+4=10, 4+1=5, 1!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2 = 1+1), row 3 is '1, 3, 3, 1' (3 = 1+2, middle 3 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! For $(x + y)^6$, row 6 is 1,6,15,20,15,6,1, and the $x^3 y^3$ term corresponds to the 4th coefficient (for $y^3$), which is 20. Choice C correctly identifies the coefficient as 20 using row 6 of Pascal's Triangle. Picking 15 (choice A) might mean choosing the adjacent coefficient—note that for equal exponents in even $n$, it's the middle one: here, 20 for $k=3$ in $C(6,3)$. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to $2^n$ (row 4 sums to 16 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(x + y)^6$ expands using row 6 of Pascal's Triangle, which is 1, 6, 15, 20, 15, 6, 1. The 4th term (counting from term 1) has the 4th coefficient (20) and follows the pattern where x-powers decrease and y-powers increase: term 1 is $x^6$, term 2 is $6x^5y$, term 3 is $15x^4y^2$, and term 4 is $20x^3y^3$. Choice B correctly identifies this as $20x^3y^3$, with coefficient 20 from Pascal's Triangle and exponents summing to 6. Choice A ($15x^2y^4$) would be the 5th term, Choice C ($15x^4y^2$) is the 3rd term, and Choice D ($6x^3y^3$) uses the wrong coefficient. Quick Pascal's Triangle construction for row 6: start with row 5 (1, 5, 10, 10, 5, 1), then build row 6 as 1, $1+5=6$, $5+10=15$, $10+10=20$, $10+5=15$, $5+1=6$, 1—systematic and reliable!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(x + y)^n$ expands to a sum of $n+1$ terms where coefficients come from row $n$ of Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For example, $(x + y)^4$ uses row 4 of Pascal's Triangle ($1, 4, 6, 4, 1$) to give: $1 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4$. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For $(x + y)^6$, row 6 is $1,6,15,20,15,6,1$, so the 4th term (starting from term 1 as $x^6$) is $20 x^3 y^3$. Choice B correctly identifies the 4th term as $20x^3 y^3$ using row 6. A mistake like $15x^3 y^3$ (choice A) could be from picking the wrong position—count carefully: 1st: $x^6$, 2nd: $6x^5 y$, 3rd: $15x^4 y^2$, 4th: $20x^3 y^3$. The expansion recipe using Pascal's Triangle: (1) Identify $n$ (the exponent on the binomial), (2) Write or construct row $n$ of Pascal's Triangle—you'll have $n+1$ numbers, (3) Create $n+1$ terms: first has coefficient 1 and is $x^n$, last has coefficient 1 and is $y^n$, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: $[1$st coefficient$] \cdot x^n + [$2nd coefficient$] \cdot x^{n-1} \cdot y + [$3rd coefficient$] \cdot x^{n-2} \cdot y^2 + \dots$ The pattern is systematic and reliable!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' (2 = 1+1), row 3 is '1, 3, 3, 1' (3 = 1+2, middle 3 = 2+1), and so on. Once you build the triangle, you have instant access to binomial coefficients! In $(x+y)^6$ using row 6 (1,6,15,20,15,6,1), the 4th term is the coefficient 20 times $x^{6-3} y^3$, or $20x^3 y^3$, matching choice B. Choice A ($15x^4 y^2$) is actually the 3rd term—remember to count starting from term 1 as the $x^6$ term. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to $2^n$ (row 4 sums to 16 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(x + y)^n$ expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For $(x + y)^5$, we use row 5 of Pascal's Triangle $(1, 5, 10, 10, 5, 1)$ to get: $1 \cdot x^5 + 5 \cdot x^4 y + 10 \cdot x^3 y^2 + 10 \cdot x^2 y^3 + 5 \cdot x y^4 + 1 \cdot y^5$. Choice A correctly shows this expansion with all six terms having the right coefficients from Pascal's Triangle and powers that decrease for x (5→0) while increasing for y (0→5). Choice B incorrectly swaps the middle coefficients (has 5 and 10 instead of 10 and 10), while Choice C uses row 4 coefficients instead of row 5. The expansion recipe using Pascal's Triangle: (1) Identify n = 5, (2) Use row 5: $(1, 5, 10, 10, 5, 1)$, (3) Create 6 terms with decreasing x-powers and increasing y-powers, (4) Write it out systematically—the pattern is beautiful and reliable!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! For $(2x + 1)^3$, we use row 3 of Pascal's Triangle (1, 3, 3, 1) but must be careful with the coefficient 2 on x. The expansion is: $$1 \cdot(2x)^3 + 3 \cdot(2x)^2 \cdot 1 + 3 \cdot(2x) \cdot 1^2 + 1 \cdot 1^3 = 8x^3 + 3 \cdot 4x^2 + 3 \cdot 2x + 1 = 8x^3 + 12x^2 + 6x + 1.$$ Choice A correctly shows this expansion with proper handling of the 2: $(2x)^3 = 8x^3$, $(2x)^2 = 4x^2$, and $2x$ stays as $2x$. Choice B incorrectly swaps the middle coefficients (6 and 12), Choice C has $6x^3$ instead of $8x^3$ (forgetting $2^3 = 8$), and Choice D is missing the constant term 1. When expanding $(ax + b)^n$, remember to raise the entire term 'ax' to each power: $(ax)^k = a^k x^k$—don't forget to compute $a^k$ for each term!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! Pascal's Triangle is built with a beautiful pattern: each row starts and ends with 1, and each interior number equals the sum of the two numbers above it in the previous row. Row 0 is just '1', row 1 is '1, 1', row 2 is '1, 2, 1' ($2 = 1+1$), row 3 is '1, 3, 3, 1' ($3 = 1+2$, middle $3 = 2+1$), and so on. Once you build the triangle, you have instant access to binomial coefficients! For $(x + y)^6$, row 6 is 1,6,15,20,15,6,1, and the $x^4 y^2$ term is for $y^2$ (3rd position), coefficient 15. Choice B correctly identifies the coefficient as 15 using row 6 of Pascal's Triangle. Choosing 20 (choice C) might be from picking for $y^3$ instead—confirm the exponents: $x^4 y^2$ needs k=2, $C(6,2)=15$. Quick Pascal's Triangle construction: write 1s down both edges. For interior numbers, add the two directly above. Example for row 4: edges are 1, then interior: 1+3=4, 3+3=6, 3+1=4, giving row '1, 4, 6, 4, 1.' Check: row sums to $2^n$ (row 4 sums to 16 = $2^4$). This build-as-you-go method means you never need to memorize rows—just construct them!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(x + y)^n$ expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For example, $(x + y)^4$ uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: $1 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^4$. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! In $(x+y)^6$ with row 6 (1,6,15,20,15,6,1), the coefficient for $x^4 y^2$ is the third one (15, for y^2), so choice B is correct. Choice C (20) is for $y^3$ instead—ensure the y-power matches the position in the row (k+1 for $y^k$). The expansion recipe using Pascal's Triangle: (1) Identify n (the exponent on the binomial), (2) Write or construct row n of Pascal's Triangle—you'll have n+1 numbers, (3) Create n+1 terms: first has coefficient 1 and is $x^n$, last has coefficient 1 and is $y^n$, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: $[1\text{st coefficient}] \cdot x^n + [2\text{nd coefficient}] \cdot x^{n-1} y + [3\text{rd coefficient}] \cdot x^{n-2} y^2 + \dots$ The pattern is systematic and reliable!

This question tests your understanding of the Binomial Theorem—a formula for expanding $(x + y)^n$ using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says $(a + b)^n$ expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, a-powers decrease from n to 0, and b-powers increase from 0 to n. For $(a + b)^4$, we use row 4 (given as $1, 4, 6, 4, 1$) to build: $1 \cdot a^4 + 4 \cdot a^3 b + 6 \cdot a^2 b^2 + 4 \cdot a b^3 + 1 \cdot b^4$. Choice A correctly shows all five terms with proper coefficients and powers summing to 4 in each term. Choice B has a missing coefficient (just ab^3 instead of 4ab^3), Choice C uses row 5 coefficients ($5, 10, 5$) instead of row 4, and Choice D has a typo with a^2b instead of a^3b in the second term. The expansion recipe using Pascal's Triangle: (1) Identify n = 4, (2) Use row 4: $1, 4, 6, 4, 1$, (3) Create 5 terms with decreasing a-powers and increasing b-powers, (4) Write systematically—this method never fails!