Miscellaneous problems and multinomial preview

Once you can write the general term, find the middle term, and use the basic identities, the binomial theorem becomes a problem-solving toolkit. Real exam problems mix several ideas at once. This subtopic collects the common combinations and previews the multinomial theorem for trinomial and higher expansions.

Definitions and reminders

Recall:

• General term: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
• Coefficient of $x^k$: solve the exponent equation for $r$, then plug in.
• Sum of coefficients of $f(x) = (a_0 + a_1 x + \dots + a_n x^n)$: evaluate $f(1)$.
• Sum of coefficients of even/odd powers: $\dfrac{f(1) \pm f(-1)}{2}$.

Two recurring problem types

Type A , ratios and equations of terms

Often a problem specifies "coefficients of $T_5$ and $T_8$ are equal" or "$T_3 : T_5 =$ some ratio". The strategy is the same: write both terms, divide, simplify.

Pattern. $\dfrac{T_{r+1}}{T_r} = \dfrac{n - r + 1}{r} \cdot \dfrac{b}{a}$. Repeated use of this recursion is often quicker than computing $T_r$ from scratch.

Type B , finding $n$ from coefficient conditions

A statement like "coefficient of $x^7$ equals coefficient of $x^8$ in $(3 + ax)^{15}$" gives an equation in $a$ (or $n$). Set up using the general term, equate, solve.

The multinomial theorem (preview)

The binomial theorem expands a sum of two terms raised to a power. The multinomial generalises to more terms: $(x_1 + x_2 + \dots + x_k)^n = \sum_{n_1 + n_2 + \dots + n_k = n} \frac{n!}{n_1!\, n_2! \cdots n_k!} \, x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}.$

For $k = 2$, this reduces to the ordinary binomial theorem since $n_1 + n_2 = n$ and $\dfrac{n!}{n_1!\, n_2!} = \binom{n}{n_1}$.

The number of distinct terms in the expansion of $(x_1 + \dots + x_k)^n$ equals the number of non-negative integer solutions to $n_1 + \dots + n_k = n$, which is $\binom{n + k - 1}{k - 1}$.

Worked examples

Example 1. Find $n$ if the coefficients of $T_2$, $T_3$, $T_4$ in $(1 + x)^n$ are in AP.

Coefficients: $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$. AP condition: $2 \binom{n}{2} = \binom{n}{1} + \binom{n}{3}$. $2 \cdot \frac{n(n-1)}{2} = n + \frac{n(n-1)(n-2)}{6}.$ $n(n - 1) = n + \frac{n(n-1)(n-2)}{6}$. Divide by $n$ (assume $n > 0$): $n - 1 = 1 + \frac{(n-1)(n-2)}{6}$. $6(n - 2) = (n-1)(n-2)$. Assuming $n \ne 2$: $6 = n - 1$, so $n = 7$.

Example 2. Find the coefficient of $x^5$ in $(1 + x + x^2)^4$.

By multinomial, $(1 + x + x^2)^4 = \sum \frac{4!}{n_1!\, n_2!\, n_3!}\, x^{n_2 + 2 n_3}, \quad n_1 + n_2 + n_3 = 4.$ Need $n_2 + 2 n_3 = 5$ with $n_1, n_2, n_3 \ge 0$ and sum $4$.

Possible $(n_3, n_2, n_1)$: $(2, 1, 1), (1, 3, 0)$.

• $(2, 1, 1)$: $\frac{4!}{1!\,1!\,2!} = 12$.
• $(1, 3, 0)$: $\frac{4!}{0!\,3!\,1!} = 4$.

Coefficient: $12 + 4 = 16$.

Example 3. The coefficient of $x^{17}$ in $(1 + x)^{50} + x(1 + x)^{49} + x^2 (1 + x)^{48} + \dots + x^{50}$.

This is a geometric series: $\sum_{k=0}^{50} x^k (1 + x)^{50 - k} = (1 + x)^{50} \cdot \frac{1 - (x/(1+x))^{51}}{1 - x/(1+x)}.$ The simpler approach: the coefficient of $x^{17}$ in $x^k (1 + x)^{50 - k}$ is $\binom{50 - k}{17 - k}$ for $0 \le k \le 17$. Summing: $\sum_{k=0}^{17} \binom{50 - k}{17 - k} = \sum_{j=0}^{17} \binom{50 - 17 + j}{j} = \sum_{j=0}^{17} \binom{33 + j}{j}.$ Using the hockey-stick identity, $\sum_{j=0}^{m} \binom{r + j}{j} = \binom{r + m + 1}{m}$, this equals $\binom{51}{17}$.

Example 4. Find the term independent of $x$ in $\left(\sqrt[3]{x} + \dfrac{1}{2 \sqrt[3]{x}}\right)^{18}$.

$T_{r+1} = \binom{18}{r} x^{(18-r)/3} \cdot \dfrac{1}{2^r} x^{-r/3} = \binom{18}{r} \dfrac{1}{2^r} x^{(18 - 2r)/3}$.

Set $(18 - 2r)/3 = 0 \Rightarrow r = 9$. $T_{10} = \binom{18}{9} \cdot \dfrac{1}{2^9} = \dfrac{48620}{512} = \dfrac{12155}{128}$.

Example 5. In the expansion of $(1 + x)^{43}$, the coefficients of $(2r + 1)$th and $(r + 2)$th terms are equal. Find $r$.

Coefficients $\binom{43}{2r}$ and $\binom{43}{r + 1}$ equal $\Rightarrow$ either $2r = r + 1$ (i.e., $r = 1$) or $2r + (r + 1) = 43$ (i.e., $r = 14$). Both valid.

Try it yourself

1. Find $n$ if coefficients of $T_5$, $T_6$, $T_7$ in $(1 + x)^n$ are in AP.
2. In $(1 + x)^n$, if coefficient of $x^9$ equals coefficient of $x^{10}$, find $n$.
3. Find the coefficient of $x^4$ in $(1 + x + x^2 + x^3)^{11}$. (Hint: $(1 + x + x^2 + x^3) = (1 + x)(1 + x^2)$.)
4. Find the coefficient of $x^6$ in $(1 + x)^5 (1 + x^2)^4$.
5. Find the term independent of $x$ in $\left(2 x^2 + \dfrac{3}{x}\right)^{15}$.
6. If $T_3 = 720$ in $(a + x)^6$ at $x = 1$, find $a$ (where $a > 0$).
7. Show that the greatest coefficient in $(1 + x)^n$ at $x = 1$ is $\binom{n}{\lfloor n/2 \rfloor}$.
8. Find the coefficient of $x^7$ in $(1 + x^2 - x^3)^8$.
9. How many terms are in the expansion of $(x + y + z)^{10}$?
10. Find the number of terms in $(x + y + z + w)^{12}$.
11. In $(1 + x)^{2n}$, prove that the coefficient of $x^n$ is the largest.
12. Find the sum of the coefficients of even powers of $x$ in $(1 + x + x^2)^{10}$.

Pitfalls / Tricks

• When two binomial coefficients are equal, $\binom{n}{a} = \binom{n}{b}$ implies $a = b$ or $a + b = n$ , don't forget the second case.
• For trinomial expansions, count terms via $\binom{n + 2}{2}$, not $n + 1$.
• Substituting $x = 1$, $x = -1$, or $x = i$ in a polynomial is a quick way to extract sums of subsets of coefficients.
• Insight. Hard problems decompose into simple steps: identify $a, b, n$; write $T_{r+1}$; impose the given condition; solve. Don't try to leap.