The Binomial Theorem: Additional Practice (Set 1)

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

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

The term is constant when $4 - 2k = 0$, so $k = 2$.

Evaluate the pieces: $\binom{4}{2} = 6$, $3^{4-2} = 9$, and $(-1)^2 = 1$.

$6 \times 9 \times 1 = 54$

The sign is positive because $k = 2$ is even.

Use row $4$ of Pascal's triangle, $1,\ 4,\ 6,\ 4,\ 1$, with $A = 1$ and $B = 0.1$. Every power of $1$ is $1$, so

$(1 + 0.1)^4 = 1 + 4(0.1) + 6(0.1)^2 + 4(0.1)^3 + (0.1)^4$

Each power of $0.1$ just shifts the decimal point: $0.01$, then $0.001$, then $0.0001$.

$1 + 0.4 + 0.06 + 0.004 + 0.0001 = 1.4641$

The digits $1, 4, 6, 4, 1$ of the answer are exactly row $4$, which is no coincidence: the powers of $0.1$ line the coefficients up in separate decimal places, and none of them carries.

Take $A = x^2$ and $B = y$. The general term is

$\binom{6}{k}(x^2)^{6-k}y^{k} = \binom{6}{k}\,x^{12-2k}\,y^{k}$

The factor $y^4$ forces $k = 4$, and then the power of $x$ is $12 - 2(4) = 4$, matching the $x^4$ asked for, so the term really is there.

$\binom{6}{4} = 15$

The value $20$ is $\binom{6}{3}$, the middle entry of row $6$, which belongs to a different term.

The terms are numbered from $k = 0$, so the second term has $k = 1$ and the third has $k = 2$. Their coefficients are $\binom{n}{1}$ and $\binom{n}{2}$.

The first one settles it immediately, since $\binom{n}{1} = n$ counts the $n$ single factors that could donate the lone $b$:

$\binom{n}{1} = n = 8$

Check against the other given value: $\binom{8}{2} = \dfrac{8 \times 7}{2} = 28$, which agrees. So $n = 8$, and row $8$ does begin $1,\ 8,\ 28$.

Write $\dfrac{2}{x}$ as $2x^{-1}$ so that the exponents can be added. The general term is

$\binom{5}{k}(x^2)^{5-k}\left(2x^{-1}\right)^{k} = \binom{5}{k}\,2^{k}\,x^{10 - 2k - k} = \binom{5}{k}\,2^{k}\,x^{10-3k}$

A lone $x$ means the exponent is $1$:

$10 - 3k = 1 \quad\Longrightarrow\quad k = 3$

With $\binom{5}{3} = 10$ and $2^3 = 8$, the coefficient is $10 \times 8 = 80$.

Choosing which $12$ of the $15$ factors donate a $y$ is the same act as choosing which $3$ donate an $x$: fix either choice and the other is decided. So the two counts must agree.

$\binom{n}{k} = \binom{n}{n-k} \quad\Longrightarrow\quad \binom{15}{12} = \binom{15}{15-12} = \binom{15}{3}$

The symmetry only reflects $k$ within the same row, so it never changes the top number. Both $\binom{15}{12}$ and $\binom{15}{3}$ equal $455$, while $\binom{15}{4} = 1365$ and $\binom{12}{3} = 220$.

First, note the range. A binomial coefficient $\binom{n}{k}$ is only defined for $k \le n$, and the larger index on offer here is $5$, so $n \ge 5$ before any algebra begins.

Write each side with the closed form and set them equal:

$\frac{n!}{3!\,(n-3)!} = \frac{n!}{5!\,(n-5)!}$

The $n!$ on top is the same on both sides, so the denominators must match: $3!\,(n-3)! = 5!\,(n-5)!$. Peel two factors off the larger factorial, using $(n-3)! = (n-3)(n-4)(n-5)!$, and the $(n-5)!$ cancels as well:

$6\,(n-3)(n-4) = 120 \quad\Longrightarrow\quad (n-3)(n-4) = 20$

Since $n \ge 5$, both $n-3$ and $n-4$ are positive, and the only two consecutive positive whole numbers with product $20$ are $5$ and $4$. So $n - 3 = 5$, giving $n = 8$.

Check it on row $8$, which is $1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1$: the entries at $k = 3$ and $k = 5$ are both $56$. The symmetry $\binom{n}{k} = \binom{n}{n-k}$ is the reason to expect a solution at all, since $3$ and $5$ are reflections of each other exactly when $3 + 5 = n$.

The bases differ, so the exponents cannot be merged. Expand each factor far enough to reach $x^2$, using rows $4$ and $3$ of the triangle:

$(1+x)^4 = 1 + 4x + 6x^2 + \cdots, \qquad (1+2x)^3 = 1 + 6x + 12x^2 + \cdots$

An $x^2$ in the product comes from any pair of terms whose powers of $x$ add to $2$, so there are three ways: the constant of the first times the $x^2$ of the second, the $x$ terms of each, and the $x^2$ of the first times the constant of the second.

$1(12) + 4(6) + 6(1) = 12 + 24 + 6 = 42$

Taking only the middle pairing gives $24$, which is why all three have to be collected.

Write both parts as powers of $x$: $\sqrt{x} = x^{1/2}$ and $\dfrac{1}{x} = x^{-1}$. The general term is then

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

A term independent of $x$ has exponent zero, so multiply through by $2$ and solve:

$\frac{9-k}{2} = k \quad\Longrightarrow\quad 9 - k = 2k \quad\Longrightarrow\quad k = 3$

Row $9$ is $1,\ 9,\ 36,\ 84,\ 126,\ 126,\ 84,\ 36,\ 9,\ 1$, so the term is $\binom{9}{3} = 84$. Rational exponents cause no trouble here, because the exponent rules are the same ones.

Row $6$ of Pascal's triangle is $1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$, so the entries at the even indices $k = 0, 2, 4, 6$ are $1$, $15$, $15$, and $1$.

Add them:

$1 + 15 + 15 + 1 = 32$

The whole row sums to $2^6 = 64$, and this is exactly half of it. That is not luck: the odd-index entries $6 + 20 + 6$ also come to $32$, because the alternating sum $(1-1)^6 = 0$ forces the two halves to be equal.

Take $A = x^2$ and $B = -2x$. Both parts carry an $x$, so collect the exponents carefully:

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

Set the exponent equal to $4$, which gives $k = 2$. Then $\binom{3}{2} = 3$ and $(-2)^2 = 4$, positive because the exponent is even.

$3 \times 4 = 12$

As a check, factoring first gives $(x^2 - 2x)^3 = x^3(x-2)^3 = x^3(x^3 - 6x^2 + 12x - 8)$, whose $x^4$ term is indeed $12x^4$.

In the general term $\binom{n}{k}x^{n-k}y^{k}$, the exponent on $y$ is $k$ and the exponent on $x$ is $n - k$. Matching the given term gives $k = 4$ and $n - k = 7$.

$n - 4 = 7 \quad\Longrightarrow\quad n = 11$

The same answer comes from the rule that every term's exponents add to $n$: $7 + 4 = 11$.