The Binomial Theorem: Practice

The general term is $\binom{5}{k}x^{5-k}3^k$. You want $x^2$, so $5 - k = 2$ and therefore $k = 3$.

Row $5$ gives $\binom{5}{3} = 10$, and the constant carries the power $3^3 = 27$:

$\binom{5}{3}\,3^3 = 10 \times 27 = 270$

Answering $10$ means forgetting that the $3$ is raised to the third power too.

For a difference, the coefficient of $a^{5-k}b^k$ is $(-1)^k\binom{5}{k}$.

The term $a^3b^2$ has $k = 2$, which is even, so $(-1)^2 = 1$ and the sign stays positive. Row $5$ gives $\binom{5}{2} = 10$.

$(-1)^2\binom{5}{2} = 10$

The minus sign only reaches the terms with an odd power of $b$, and this is not one of them.

The constant term is the one with no $x$ left, which happens when every factor donates its $2$. That is $k = 5$.

The general term is $\binom{5}{k}x^{5-k}2^k$, so at $k = 5$ it is $\binom{5}{5}x^0 2^5$, and $\binom{5}{5} = 1$.

$1 \cdot 2^5 = 32$

You can also get it by putting $x = 0$ into $(x+2)^5$, which leaves $2^5 = 32$.

Use the symmetry $\binom{n}{k} = \binom{n}{n-k}$, which holds because choosing the $7$ factors that donate a $y$ is the same act as choosing the $3$ that donate an $x$.

$\binom{10}{7} = \binom{10}{3}$

Row $10$ of Pascal's triangle is $1,\ 10,\ 45,\ 120,\ 210,\ 252,\ 210,\ 120,\ 45,\ 10,\ 1$, and the entry at $k = 3$ is $120$.

$\binom{10}{7} = 120$

The expansion is an identity, true for every value of $x$, so you may substitute whatever is convenient. Setting $x = 1$ makes every power of $x$ equal to $1$, which leaves exactly the sum of the coefficients.

So the sum of the coefficients is the value of the unexpanded expression at $x = 1$:

$(2 \cdot 1 + 3)^4 = 5^4 = 625$

No expansion needed. As a check, $(2x+3)^4 = 16x^4 + 96x^3 + 216x^2 + 216x + 81$, and $16 + 96 + 216 + 216 + 81 = 625$.

Take $A = x$ and $B = -2y$. The general term is $\binom{4}{k}x^{4-k}(-2y)^k$, and $y^2$ needs $k = 2$.

Expand the whole of $-2y$ to the second power: $(-2y)^2 = 4y^2$, positive because the exponent is even. Row $4$ gives $\binom{4}{2} = 6$.

$\binom{4}{2}\,(-2)^2 = 6 \times 4 = 24$

The coefficient is $24$. The minus sign disappears here because $k$ is even, and the $2$ must be squared, not merely copied.

The number of terms depends only on the exponent, not on the coefficients inside the binomial. There is one term for each whole number $k$ from $0$ to $11$.

$11 + 1 = 12$

The messy coefficients $2$ and $-5$ change the numbers in front of each term, but never how many terms there are.

Row $10$ of Pascal's triangle is $1,\ 10,\ 45,\ 120,\ 210,\ 252,\ 210,\ 120,\ 45,\ 10,\ 1$. The entries rise to the middle and then fall away symmetrically.

The single largest entry is the middle one, at $k = 5$:

$\binom{10}{5} = 252$

That is the coefficient of $x^{10-5}y^5 = x^5y^5$. The coefficient of $x^6y^4$ is $\binom{10}{4} = 210$, which is smaller.

The term takes a $y$ from $6$ of the $8$ factors, so $k = 6$ and the coefficient is $\binom{8}{6}$.

By symmetry $\binom{8}{6} = \binom{8}{2}$, and row $8$ of the triangle is $1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1$.

$\binom{8}{6} = 28$

The entry $70$ is the middle one, $\binom{8}{4}$, and belongs to $x^4y^4$.

Take $A = 1$ and $B = -x$. Since every power of $1$ is $1$, the general term is $\binom{5}{k}(-x)^k = (-1)^k\binom{5}{k}x^k$.

Row $5$ is $1,\ 5,\ 10,\ 10,\ 5,\ 1$, and the factor $(-1)^k$ flips the sign on the odd powers of $x$:

$(1 - x)^5 = 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5$

The last term has $k = 5$, which is odd, so it must be $-x^5$, not $+x^5$.

The coefficient of $a^{7-k}b^k$ is $(-1)^k\binom{7}{k}$, and $\binom{7}{k}$ is a count, so it is always positive. The sign therefore comes from $(-1)^k$ alone, and the coefficient is negative exactly when $k$ is odd.

List the odd values of $k$ between $0$ and $7$: $k = 1, 3, 5, 7$.

$\text{four odd values of } k \;\Longrightarrow\; \text{four negative coefficients}$

The signs run $+, -, +, -, +, -, +, -$ across the eight terms, so exactly half are negative.

Pascal's rule says an entry equals the sum of the two entries above it:

$\binom{10}{5} = \binom{9}{4} + \binom{9}{5}$

Substitute the two given values:

$\binom{10}{5} = 126 + 126 = 252$

This is the middle, and therefore the largest, entry of row $10$.