The Binomial Theorem: Practice

Write $\dfrac{1}{x}$ as $x^{-1}$ so the powers can be added. The general term is

$\binom{8}{k}x^{8-k}\left(x^{-1}\right)^{k} = \binom{8}{k}\,x^{8-2k}$

A constant term has no $x$ left, so set the exponent to zero:

$8 - 2k = 0 \quad\Longrightarrow\quad k = 4$

Row $8$ of Pascal's triangle is $1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1$, so the constant term is $\binom{8}{4} = 70$.

Take $A = x^2$ and $B = -x^{-1}$. The general term is

$\binom{7}{k}(x^2)^{7-k}\left(-x^{-1}\right)^{k} = \binom{7}{k}(-1)^k\,x^{14 - 2k - k} = \binom{7}{k}(-1)^k\,x^{14-3k}$

Set the exponent equal to the target and solve for $k$:

$14 - 3k = 5 \quad\Longrightarrow\quad k = 3$

Row $7$ gives $\binom{7}{3} = 35$, and $k = 3$ is odd, so $(-1)^3 = -1$. The coefficient is $-35$.

The expansion has $6 + 1 = 7$ terms, so the middle one is the fourth, which has $k = 3$.

With $A = 2x$ and $B = -1$, the term is

$\binom{6}{3}(2x)^{3}(-1)^{3} = 20 \cdot 8x^3 \cdot (-1)$

Evaluate: $\binom{6}{3} = 20$, $(2x)^3 = 8x^3$, and $(-1)^3 = -1$.

$-160x^3$

Answering $-20x^3$ means cubing only the $x$ and leaving the $2$ behind.

The expansion is an identity, so substitute the values that turn every power of a variable into $1$. Setting $x = 1$ and $y = 1$ leaves exactly the sum of the coefficients.

$\big(3(1) - 2(1)\big)^5 = (3 - 2)^5 = 1^5 = 1$

The sum is $1$. The coefficients themselves are large and alternate in sign, and they cancel almost completely, which is why no expansion is worth doing here.

Cancel $8!$ against the tail of $12!$, which leaves four factors on top and $4!$ underneath:

$\binom{12}{4} = \frac{12 \times 11 \times 10 \times 9 \times 8!}{(4 \times 3 \times 2 \times 1) \times 8!} = \frac{12 \times 11 \times 10 \times 9}{24}$

Cancel before multiplying: $12 \div 24 = \tfrac{1}{2}$, and $10 \div 2 = 5$, so the product is $11 \times 5 \times 9$.

$\binom{12}{4} = 495$

The value $11880$ is the numerator alone, with the $4!$ never divided out.

Do not expand the two factors separately. The exponent rules combine them first:

$(1 + x)^3(1 + x)^4 = (1 + x)^{3+4} = (1 + x)^7$

In $(1 + x)^7$ the coefficient of $x^k$ is simply $\binom{7}{k}$, so you want $\binom{7}{2}$. Row $7$ is $1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1$.

$\binom{7}{2} = 21$

The value $12$ is the product of the two $x$-coefficients, $3$ from $(1+x)^3$ and $4$ from $(1+x)^4$. That is only one of the three ways a product can build an $x^2$ term, and merging the exponents first saves you from having to chase the other two.

The general term is $\binom{5}{k}x^{5-k}a^{k}$, and $x^4$ needs $k = 1$. That term is $\binom{5}{1}x^4a = 5ax^4$.

Set its coefficient equal to the given value:

$5a = 20 \quad\Longrightarrow\quad a = 4$

Checking, $(x + 4)^5$ has second term $5x^4(4) = 20x^4$. Answering $20$ ignores the $\binom{5}{1} = 5$ that multiplies $a$.

Take $A = x^2$ and $B = -2x^{-1}$. The general term is

$\binom{9}{k}(x^2)^{9-k}\left(-2x^{-1}\right)^{k} = \binom{9}{k}(-2)^{k}\,x^{18 - 2k - k} = \binom{9}{k}(-2)^{k}\,x^{18-3k}$

The term is constant when $18 - 3k = 0$, so $k = 6$.

Row $9$ is $1,\ 9,\ 36,\ 84,\ 126,\ 126,\ 84,\ 36,\ 9,\ 1$, giving $\binom{9}{6} = 84$, and $(-2)^6 = 64$ is positive because the exponent is even.

$84 \times 64 = 5376$

Choosing which $3$ of the $9$ factors donate a $y$ is the same act as choosing which $6$ donate an $x$, so the two counts agree. In symbols, $\binom{n}{k} = \binom{n}{n-k}$.

$\binom{9}{3} = \binom{9}{9-3} = \binom{9}{6}$

Row $9$ is $1,\ 9,\ 36,\ 84,\ 126,\ 126,\ 84,\ 36,\ 9,\ 1$, and the value $84$ appears at $k = 3$ and $k = 6$ and nowhere else, since the entries rise strictly to the middle and then fall.

Substitute $x = y = 1$ into the Binomial Theorem. The left side becomes $(1 + 1)^7$, and every power of $x$ and $y$ on the right becomes $1$, leaving only the coefficients.

$\sum_{k=0}^{7}\binom{7}{k} = (1+1)^7 = 2^7 = 128$

The same count reads directly off the definition: each of the $7$ factors independently donates an $x$ or a $y$, giving $2^7$ picks in all, and sorting them by the number of y's is exactly what the sum does. Adding row $7$ by hand confirms it: $1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = 128$.

Take $A = 2x$ and $B = -y$. The general term is $\binom{8}{k}(2x)^{8-k}(-y)^{k}$, and $y^3$ needs $k = 3$, which leaves $x^{8-3} = x^5$ as required.

Evaluate the three pieces: $\binom{8}{3} = 56$, $(2x)^5 = 32x^5$, and $(-y)^3 = -y^3$.

$56 \times 32 \times (-1) = -1792$

The $2$ is raised to the fifth power, not merely copied, and the odd $k$ makes the sign negative.

An expansion of $(x + y)^n$ has $n + 1$ terms, one for each $k$ from $0$ to $n$. So $n + 1 = 15$ gives $n = 14$, and the exponents check out, since $12 + 2 = 14$.

The coefficient of $x^{12}y^2$ is $\binom{14}{2}$, which the closed form settles at once:

$\binom{14}{2} = \frac{14 \times 13}{2} = 91$

Using $n = 15$ instead would give $\binom{15}{2} = 105$, the classic off-by-one here.