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1. Algebra — Equations, Progressions and Series

Binomial theorem

The binomial theorem provides an algebraic expansion formula for powers of binomial expressions, forming the foundation of combinatorial identities and series approximations. This topic covers general terms, middle terms, and coefficient properties.

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Definition and statement

Binomial Theorem for Positive Integral Index
For any positive integer $n$, $$(x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n}y^n$$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ are the binomial coefficients.

📐 Geometric Interpretation

Each term in the expansion represents the volume contribution of sub-hypercubes in an $n$-dimensional space divided according to lengths $x$ and $y$.

⚙️ Algebraic & Structural Meaning

Establishes a deep bridge between combinatorics and polynomial algebra, facilitating rapid evaluation of sums involving combinations and powers.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Index $n \in \mathbb{N} \cup {0}$ Discrete set of integers Strictly Mandatory for finite expansion If $n$ is fractional or negative, the series becomes infinite and requires $|x/y| < 1$ for convergence.
Commutativity of ring elements $x, y$ General algebra context ($xy = yx$) Strictly Mandatory If $xy \neq yx$ (non-commutative algebra, e.g., matrices), standard binomial expansion fails without using the full Leibniz rule.
Convergence condition for general index $|x| < |y|$ Essential for infinite binomial series Divergence of the series if $|x| > |y|$.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

General Term Extraction

$$T_{r+1} = \binom{n}{r} x^{n-r} y^r$$

Rule: To find the coefficient of $x^m$ in $(ax^p + bx^{-q})^n$, equate the index of $x$ in $T_{r+1}$ to $m$ to find $r$ instantly.

Application: undefined
02

Sum of Binomial Coefficients

$$ \sum_{r=0}^{n} \binom{n}{r} = 2^n, \quad \sum_{r=0}^{\lfloor n/2 \rfloor} \binom{n}{2r} = 2^{n-1} $$

Rule: Sum of odd coefficients equals sum of even coefficients, both equaling $2^{n-1}$.

Application: undefined
03

Greatest Binomial Coefficient

\text{If } n \text{ is even, } \binom{n}{n/2}; \text{ if } n \text{ is odd, } \binom{n}{(n-1)/2} = \binom{n}{(n+1)/2}

Rule: The middle term possesses the maximum binomial coefficient.

Application: undefined
04

Ratio of Consecutive Coefficients

$$ \frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} \frac{y}{x} $$

Rule: Use this ratio directly to find the numerically greatest term in a binomial expansion.

Application: undefined
05

Multinomial Theorem Extension

$$ (x_1 + x_2 + \dots + \k)^n = \sum \frac{n!}{n_1! n_2! \dots n_k!} x_1^{n_1} x_2^{n_2} \dots x_k^{n_k} $$

Rule: Total number of terms is $\binom{n+k-1}{k-1}$.

Application: undefined
06

Remainder Finding via Binomial

$$ (1 + x)^n = 1 + nx + \dots $$

Rule: Express large powers like $(9)^{20}$ as $(10 - 1)^{20}$ to isolate remainders modulo $10$ or other integers.

Application: undefined

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing $T_{r}$ with $T_{r+1}$

Common Pitfall: Taking the $r$-th term formula as $\binom{n}{r}x^{n-r}y^r$ instead of the correct $(r+1)$-th term $T_{r+1}$.

Counterexample: For $T_5$, $r=4$, not $5$.

Why it fails: Off-by-one error leads to wrong coefficient extraction.

🛡️ Exam Defense: Always write $T_{r+1} = \dots$ and explicitly identify $r$ as the lower index of the binomial coefficient.
⚠️ COMMON ERROR #02

Ignoring Signs in Binomial Expansion

Common Pitfall: Forgetting that $(x - y)^n$ has alternating signs: $(-1)^r$ must be attached to $y^r$.

Counterexample: (x - 2)^n \neq \sum \binom{n}{r} x^{n-r} 2^r

Why it fails: Misses the alternating negative signs, leading to incorrect sum evaluations.

🛡️ Exam Defense: Treat the binomial as $(x + (-y))^n$ whenever a negative sign is present.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Positive Integral Index$n \in \mathbb{N}$$(x+y)^n = \sum \binom{n}{r}x^{n-r}y^r$Finite sum of combinatorial volumesKeywords: 'Coefficient of $x^k$', 'independent term'
Negative / Fractional Index$n \in \mathbb{Q} \setminus \mathbb{N}$, $|x/y| < 1$$(1+x)^{-n} = 1 - nx + \frac{n(n+1)}{2!}x^2 - \dots$Infinite convergent power seriesKeywords: 'Infinite series sum', 'approximations'
Multinomial Theorem$k$ variables, $n \in \mathbb{N}$Total terms = $\binom{n+k-1}{k-1}$Partition of integer $n$ into $k$ partsKeywords: 'Total number of terms in expansion'
Exponential / Logarithmic Series$x \in \mathbb{R}$$e^x = 1 + x + \frac{x^2}{2!} + \dots$Limiting case of binomial expansionsKeywords: 'Sum of series', 'exponential expansion'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the coefficient of $x^7$ in the expansion of $\left(ax^2 + \frac{1}{bx}\right)^{11}$.

(A) $\binom{11}{5} a^3 b^{-6}$(B) $\binom{11}{4} a^4 b^{-7}$(C) $\binom{11}{6} a^3 b^{-5}$(D) $\binom{11}{5} a^6 b^{-5}$
✓ Correct Answer: (D)
📚 Method A: Formal Textbook Derivation
Step 1: Write $T_{r+1} = \binom{11}{r} (ax^2)^{11-r} (b^{-1} x^{-1})^r = \binom{11}{r} a^{11-r} b^{-r} x^{22 - 3r}$. Step 2: Equate power $22 - 3r = 7 \implies 3r = 15 \implies r = 5$. Step 3: Substitute $r=5$ to get coefficient $\binom{11}{5} a^6 b^{-5}$.
⚡ Method B: 25-Second Exam Shortcut
Set $22 - 3r = 7 \implies r=5$ instantly. Match power of $a$ as $11-5=6$ and $b$ as $-5$.
💡 Tip: Check powers carefully; variable indices must be resolved without arithmetic slips.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the coefficients of the $(2r+1)$-th and $(r+2)$-th terms in $(1+x)^{43}$ are equal, find $r$.

(A) $14$(B) $12$(C) $10$(D) $15$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify terms: $T_{2r+1} = \binom{43}{2r}$ and $T_{r+2} = \binom{43}{r+1}$. Step 2: Equate coefficients $\binom{43}{2r} = \binom{43}{r+1}$. Step 3: Use property $\binom{n}{a} = \binom{n}{b} \implies a=b$ or $a+b=n$. Here $2r + r + 1 = 43 \implies 3r = 42 \implies r = 14$.
⚡ Method B: 25-Second Exam Shortcut
Instantly apply $2r + (r+1) = 43 \implies 3r = 42 \implies r=14$.
💡 Tip: Always test both cases: $a=b$ gives $r=1$ (rejected if out of bounds), $a+b=n$ gives the valid solution.

Practice questions

Choose an answer to check it and see the solution:

Q1
The total number of terms in the expansion of $(x + y + z)^{n}$ is:
A $\frac{n(n+1)}{2}$
B $\frac{(n+1)(n+2)}{2}$
C n(n+1)
D $\frac{n(n-1)}{2}$
Derivation & Analysis:

By the multinomial formula, the number of terms for $k=3$ variables is $\binom{n+3-1}{3-1} = \binom{n+2}{2} = \frac{(n+1)(n+2)}{2}$.

Q2
If $\sum_{r=0}^{n} \binom{n}{r} = 4096$, then the value of $n$ is:
A 10
B 11
C 12
D 13
Derivation & Analysis:

We know $\sum_{r=0}^{n} \binom{n}{r} = 2^n$. Thus $2^n = 4096 = 2^{12} \implies n = 12$.

Q3
The term independent of $x$ in $\left(x - \frac{2}{x^2}\right)^{15}$ is:
A $-\binom{15}{5} 2^5$
B $\binom{15}{5} 2^5$
C $-\binom{15}{6} 2^6$
D $\binom{15}{6} 2^6$
Derivation & Analysis:

General term $T_{r+1} = \binom{15}{r} x^{15-r} (-2)^r x^{-2r} = \binom{15}{r} (-2)^r x^{15-3r}$. Setting $15-3r = 0 \implies r = 5$. Thus $T_6 = \binom{15}{5} (-2)^5 = -\binom{15}{5} 2^5$.

Q4
The coefficient of $x^n$ in $(1 + x)^{2n}$ is:
A $\frac{(2n)!}{(n!)^2}$
B $\frac{2n!}{n!}$
C $2^n \frac{n!}{(2n)!}$
D $\binom{2n-1}{n}$
Derivation & Analysis:

The general term in $(1+x)^{2n}$ is $\binom{2n}{r} x^r$. For $r=n$, the coefficient is $\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2}$.

Q5
If the coefficients of $2nd$, $3rd$, and $4th$ terms in the expansion of $(1+x)^n$ are in A.P., then $n$ equals:
A 7
B 14
C 6
D 11
Derivation & Analysis:

Coefficients are $\binom{n}{1}, \binom{n}{2}, \binom{n}{3}$. Since they are in A.P., $2\binom{n}{2} = \binom{n}{1} + \binom{n}{3}$. Solving this yields $n = 7$ or $n = 14$ (standard result, $n=14$ is valid).

Q6
The greatest term in the expansion of $(1 + x)^{20}$ when $x = 2$ is:
A $T_{13}$
B $T_{14}$
C $T_{15}$
D $T_{12}$
Derivation & Analysis:

Using $\frac{T_{r+1}}{T_r} = \frac{n-r+1}{r} x \ge 1 \implies \frac{21-r}{r}(2) \ge 1 \implies 42 - 2r \ge r \implies 3r \le 42 \implies r \le 14$. Thus $T_{14}$ is the greatest term.

Q7
The value of $\binom{n}{0} + 2\binom{n}{1} + 3\binom{n}{2} + \dots + (n+1)\binom{n}{n}$ is:
A $(n+1)2^{n-1}$
B $(n+2)2^{n-1}$
C $(n+1)2^n$
D $n 2^{n-1}$
Derivation & Analysis:

Rewrite general term as $(r+1)\binom{n}{r} = r\binom{n}{r} + \binom{n}{r} = n\binom{n-1}{r-1} + \binom{n}{r}$. Summing gives $n 2^{n-1} + 2^n = (n+2)2^{n-1}$.

Q8
Remainder when $2^{50}$ is divided by $7$ is:
A 1
B 2
C 4
D 6
Derivation & Analysis:

$2^3 = 8 = 7 + 1$. Thus $2^{50} = 2^2 \cdot (2^3)^{16} = 4 \cdot (7+1)^{16}$. Expanding $(7+1)^{16}$, all terms except the last are divisible by $7$. Remainder is $4 \times 1 = 4$.

Q9
If $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \dots + a_{2n} x^{2n}$, then $a_0 + a_1 + a_2 + \dots + a_{2n}$ is:
A $2^n$
B $3^n$
C $3^{2n}$
D 1
Derivation & Analysis:

Substitute $x = 1$ in the identity: $(1 + 1 + 1)^n = a_0 + a_1 + \dots + a_{2n} \implies 3^n = \sum a_i$.

Q10
The sum $\binom{n}{0} \binom{n}{r} + \binom{n}{1} \binom{n}{r+1} + \dots + \binom{n}{n-r} \binom{n}{n}$ is equal to:
A $\binom{2n}{n-r}$
B $\binom{2n}{n+r}$
C $\binom{2n}{r}$
D Both (1) and (2)
Derivation & Analysis:

By Vandermonde's Identity, this sum equals $\binom{2n}{n-r}$, which is also equal to $\binom{2n}{n+r}$ due to symmetry.

Common questions

When $n$ is even, there is only one middle term ($T_{n/2 + 1}$), but when $n$ is odd, there are two middle terms ($T_{(n+1)/2}$ and $T_{(n+3)/2}$). Miscounting $n$ parity leads to incorrect term selection.

Binomial coefficients deal with powers of two terms $(x+y)^n$, whereas multinomial coefficients handle sums of three or more terms. Their sums satisfy different combinatorial identities.

Never expand the full polynomial. Directly write down the general term $T_{r+1}$, collect all exponents of the variable, equate the exponent to the required power, and solve for $r$.

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