The Binomial Theorem: Additional Practice (Set 2)

Take $A = 2x$ and $B = -3x^{-1}$. Keeping every coefficient inside its power, the general term is

$\binom{6}{k}(2x)^{6-k}\left(-3x^{-1}\right)^{k} = \binom{6}{k}\,2^{6-k}(-3)^{k}\,x^{6-2k}$

The term is constant when $6 - 2k = 0$, so $k = 3$.

Now evaluate all three pieces: $\binom{6}{3} = 20$, $2^{6-3} = 8$, and $(-3)^3 = -27$.

$20 \times 8 \times (-27) = -4320$

The sign is negative because $k = 3$ is odd.

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

Evaluate all three pieces, raising the whole of $-3x$ to the third power: $\binom{5}{3} = 10$, $2^{5-3} = 4$, and $(-3)^3 = -27$.

$10 \times 4 \times (-27) = -1080$

The $2$ still carries a power here, which is what distinguishes this from the $(1 + Bx)^n$ case.

With $A = 1$ and $B = 2x$, every power of $A$ is $1$, so the general term is $\binom{10}{k}(2x)^k = \binom{10}{k}2^k x^k$ and the power of $x$ is simply $k$. So $x^4$ needs $k = 4$.

Row $10$ is $1,\ 10,\ 45,\ 120,\ 210,\ 252,\ 210,\ 120,\ 45,\ 10,\ 1$, giving $\binom{10}{4} = 210$, and the $2$ is raised to the fourth power along with the $x$.

$\binom{10}{4}\,2^4 = 210 \times 16 = 3360$

Stopping at $210$ forgets the $2^4$.

Setting every variable to $1$ turns each power of $x$ and $y$ into $1$, so the expansion collapses to the sum of its coefficients, while the unexpanded side stays trivial to evaluate.

$\big(1 - 2(1)\big)^7 = (-1)^7 = -1$

The sum is $-1$. The value $-128$ would be $(-2)^7$, which is what you get by mistakenly dropping the $x$ rather than setting it to $1$.

Use the closed form with $k = 2$, cancelling $(n-2)!$ against the tail of $n!$:

$\binom{n}{2} = \frac{n!}{2!\,(n-2)!} = \frac{n(n-1)}{2}$

Set that equal to $45$ and clear the fraction:

$n(n-1) = 90$

Two consecutive whole numbers multiplying to $90$ must be $10$ and $9$, so $n = 10$. (The negative root of $n^2 - n - 90 = 0$ is $n = -9$, which is not a possible number of factors.)

The general term is $\binom{8}{k}x^{8-k}3^{k}$. You want $x^5$, so $8 - k = 5$ and therefore $k = 3$.

Row $8$ is $1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1$, giving $\binom{8}{3} = 56$, and the constant carries $3^3 = 27$.

$56 \times 27 = 1512$

A common slip is to use $k = 5$, which would give $\binom{8}{5}3^5$ and the wrong term entirely; $k$ counts the factors donating the $3$, not the $x$.

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

Evaluate the three pieces: $\binom{5}{3} = 10$, $(2a)^2 = 4a^2$, and $(3b)^3 = 27b^3$.

$10 \times 4 \times 27 = 1080$

Both numerical coefficients have to be raised to their own powers, which is where $4$ and $27$ come from.

The entry is $\binom{12}{3}$, and twelve rows of addition is more work than the closed form. Cancel $9!$ against the tail of $12!$:

$\binom{12}{3} = \frac{12!}{3!\,9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6}$

Dividing gives the entry:

$\binom{12}{3} = 220$

Row $12$ runs $1,\ 12,\ 66,\ 220,\ 495,\ \ldots$, so $66$ sits at $k = 2$ and $495$ at $k = 4$.

Take $A = x^3$ and $B = 2x^{-1}$. The general term collects into a single power of $x$:

$\binom{6}{k}(x^3)^{6-k}\left(2x^{-1}\right)^{k} = \binom{6}{k}\,2^{k}\,x^{18 - 3k - k} = \binom{6}{k}\,2^{k}\,x^{18-4k}$

Set the exponent equal to $6$:

$18 - 4k = 6 \quad\Longrightarrow\quad k = 3$

With $\binom{6}{3} = 20$ and $2^3 = 8$, the coefficient is $20 \times 8 = 160$.

Take $A = \dfrac{a}{2}$ and $B = 2b$. Fractions behave like any other coefficient: the whole of $A$ gets raised to its power. The factor $b^2$ needs $k = 2$, which leaves $a^{6-2} = a^4$ as required.

$\binom{6}{2}\left(\frac{a}{2}\right)^{4}(2b)^{2} = 15 \cdot \frac{a^4}{16} \cdot 4b^2$

Now combine the numbers, since $\left(\tfrac{1}{2}\right)^4 = \tfrac{1}{16}$ and $2^2 = 4$:

$15 \times \frac{1}{16} \times 4 = \frac{60}{16} = \frac{15}{4}$

Forgetting the $2^2$ from the $2b$ would leave $\tfrac{15}{16}$.

With $A = 1$ and $B = x$, the coefficient of $x^k$ is just $\binom{n}{k}$. So the condition reads $\binom{n}{2} = 3\binom{n}{1}$, and the closed form turns it into an equation in $n$:

$\frac{n(n-1)}{2} = 3n$

Since $n \ge 1$, you may divide both sides by $n$, which leaves

$\frac{n-1}{2} = 3 \quad\Longrightarrow\quad n = 7$

Check it against row $7$: the coefficient of $x^2$ is $\binom{7}{2} = 21$ and the coefficient of $x$ is $\binom{7}{1} = 7$, and $21 = 3 \times 7$.

With $A = 2$ and $B = x$, the general term is $\binom{n}{k}2^{n-k}x^{k}$, so the two coefficients in question are $\binom{n}{2}2^{n-2}$ and $\binom{n}{3}2^{n-3}$. Setting them equal and dividing both sides by $2^{n-3}$ leaves

$2\binom{n}{2} = \binom{n}{3}$

Now use the closed form on each side:

$2 \cdot \frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6}$

For $n \ge 2$ the product $n(n-1)$ is not zero, so divide it out of both sides:

$1 = \frac{n-2}{6} \quad\Longrightarrow\quad n = 8$

Check it: $\binom{8}{2}2^6 = 28 \times 64 = 1792$ and $\binom{8}{3}2^5 = 56 \times 32 = 1792$, which agree.