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Binomial Theorem
Binomial Expansion for Positive Integer Index
The Binomial Theorem expands $(a+x)^n$ into a sum of $n+1$ terms, where each term's coefficient is a binomial coefficient from Pascal's Triangle.
$$(a+x)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} x^r$$
Each term combines a binomial coefficient with decreasing powers of $a$ and increasing powers of $x$.
$\binom{n}{r}$=Binomial coefficient — number of ways to choose $r$ items from $n$
$a^{n-r}$=First term raised to a decreasing power
$x^r$=Second term raised to an increasing power
$a = 1$
→$(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$
$x$ replaced by $-x$
→Signs alternate: $(a-x)^n = \binom{n}{0}a^n - \binom{n}{1}a^{n-1}x + \binom{n}{2}a^{n-2}x^2 - \dots$
Term Count: The expansion of $(a+x)^n$ always has exactly $n+1$ terms.
Exponent Sum Rule: In every term, the exponents of $a$ and $x$ always add up to $n$.
Coefficient Symmetry: $\binom{n}{r} = \binom{n}{n-r}$, so the first and last coefficients match, the second and second-to-last match, etc.
Pascal's Identity: $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$ — this is how each entry in Pascal's Triangle is formed from the row above.
The General Term ($T_{r+1}$)
The General Term formula gives any specific term in an expansion directly, without expanding the entire binomial.
$$T_{r+1} = \binom{n}{r} a^{n-r} x^r$$
The $(r+1)$-th term of $(a+x)^n$, counting from the beginning.
$T_{r+1}$=The term at position $r+1$ in the expansion
$\binom{n}{r}$=Binomial coefficient $\frac{n!}{r!(n-r)!}$
Finding the $k$-th term from the end
→It equals the $(n - k + 2)$-th term from the beginning
Index Rule: The $r$ value is always one less than the term number. For the 5th term, use $r = 4$.
Term from End: The $k$-th term from the end in $(a+x)^n$ is the $(n-k+2)$-th term from the beginning.
Coefficient vs Term: The 'coefficient of $x^p$' means only the numerical factor — strip out all powers of $x$ after substituting $r$.
Finding Specific Terms
A term is 'independent of $x$' (or a constant term) when the combined power of $x$ in the General Term equals zero.
Step 1: Write $T_{r+1}$ for the binomial, keeping all $x$-terms explicit.
Step 2: Consolidate all $x$ factors into a single power $x^{f(r)}$ where $f(r)$ is an expression in $r$.
Step 3: Set $f(r) = 0$ and solve for $r$. If $r$ is not a non-negative integer $\leq n$, no independent term exists.
Step 4: Substitute the integer $r$ back to compute the coefficient.
Finding term involving $x^p$: Same process but set $f(r) = p$ instead of $0$.
The number of middle terms in an expansion depends on whether $n$ is even or odd.
Even $n$: Exactly one middle term at position $\left(\frac{n}{2} + 1\right)$. For $(a+x)^{12}$, the middle term is $T_7$.
Odd $n$: Two middle terms at positions $\frac{n+1}{2}$ and $\frac{n+3}{2}$. For $(a+x)^{11}$, middle terms are $T_6$ and $T_7$.
Quick Check: Total terms $= n+1$. If odd (even $n$), one middle; if even (odd $n$), two middles.
Coefficient Sum Properties
Substituting specific values of $a$ and $x$ into the expansion produces powerful identities about binomial coefficients.
$$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$$
Setting $a = 1, x = 1$ in $(a+x)^n$ gives the sum of all binomial coefficients.
$2^n$=Total sum of all binomial coefficients for index $n$
$a = 1, x = -1$
→Alternating sum $= 0$: $\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots = 0$
Sum of All Coefficients: Put $a=1, x=1$: sum $= 2^n$.
Alternating Sum: Put $a=1, x=-1$: alternating sum $= 0$.
Even-Position = Odd-Position: $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}$.
Weighted Sum: $\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} = n \cdot 2^{n-1}$.
Binomial Series (Fractional and Negative Index)
When the index $n$ is a negative integer or a fraction, the Binomial Series gives an infinite expansion valid only when $|x| < 1$.
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$$
An infinite series that converges when $|x| < 1$. Unlike the positive integer case, it does not terminate.
$n$=The index — can be negative or fractional (e.g., $-1$, $\frac{1}{2}$, $-\frac{2}{3}$)
$|x| < 1$=Convergence condition — the series is only valid when the absolute value of $x$ is less than 1
$n = -1$
→$(1+x)^{-1} = 1 - x + x^2 - x^3 + \dots$
$n = -2$
→$(1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots$
$n = \frac{1}{2}$
→$(1+x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \dots$
Key Difference: For positive integer $n$, the expansion is finite ($n+1$ terms). For fractional/negative $n$, it is infinite.
$\binom{n}{r}$ Not Valid: The standard $\binom{n}{r}$ notation is meaningless for non-positive-integer $n$. Use the product form $\frac{n(n-1)(n-2)\dots(n-r+1)}{r!}$ instead.
Convergence Condition: The base must be written as $(1 + \text{something})^n$ where $|\text{something}| < 1$.
Factoring Technique: To expand $(a + bx)^n$ where $a \neq 1$, factor out $a^n$: $(a+bx)^n = a^n\left(1 + \frac{bx}{a}\right)^n$, then apply the series to $\left(1 + \frac{bx}{a}\right)^n$.
$(1-x)^{-1}$ Series: $(1-x)^{-1} = 1 + x + x^2 + x^3 + \dots$ — this is the geometric series.
The general term of the binomial series uses descending products instead of factorials.
$$T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!} x^r$$
Negative Index Pattern: For $(1+x)^{-m}$, the general term is $(-1)^r \binom{m+r-1}{r} x^r$.
Sign Pattern: When $n < 0$, the factors in the numerator alternate sign, producing an alternating series.
Approximation: Since $|x| < 1$ and higher powers shrink, often the first 3–4 terms give sufficient accuracy to 3 decimal places.
Applications: Approximation and Series Summation
The binomial series is used to approximate roots and powers of numbers close to perfect powers.
Strategy: Express the number as $(\text{perfect power})(1 \pm \text{small quantity})^n$, then expand and keep first few terms.
Example Pattern: $\sqrt[3]{30} = (27+3)^{1/3} = 3(1+\frac{1}{9})^{1/3} \approx 3\left(1 + \frac{1}{27} - \frac{1}{729}\right) \approx 3.107$.
Neglecting Higher Powers: When $x$ is very small, $(1+x)^n \approx 1 + nx$ (first-order approximation).
Second-Order: $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ when cube and higher powers are negligible.
Infinite series can be identified as binomial expansions by comparing terms with the standard binomial series pattern.
Method: Compare the given series with $1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$ to find $n$ and $x$.
Step 1: Set $nx = $ (second term of series) to get a relation between $n$ and $x$.
Step 2: Set $\frac{n(n-1)}{2!}x^2 = $ (third term) and solve simultaneously for $n$ and $x$.
Step 3: The sum of the series is $(1+x)^n$ with the found values of $n$ and $x$.
Application Strategy
Worked Examples
Problem Statement
Expand $(x + 2)^4$ using the Binomial Theorem.
Binomial Expansion FormulaCombination Calculation
Strategic Toolkit
$$(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$
Step 1
Identify Parameters
Set $n=4$, $a=x$, $b=2$.
$n=4, a=x, b=2$
Step 2
Write All Terms
Apply the formula for $r=0,1,2,3,4$ (giving 5 terms).
$\binom{4}{0}x^4 \cdot 1 + \binom{4}{1}x^3 \cdot 2 + \binom{4}{2}x^2 \cdot 4 + \binom{4}{3}x \cdot 8 + \binom{4}{4} \cdot 16$
Step 3
Evaluate Coefficients
Row 4 of Pascal's Triangle: $1, 4, 6, 4, 1$.
$1 \cdot x^4 + 4 \cdot 2x^3 + 6 \cdot 4x^2 + 4 \cdot 8x + 1 \cdot 16$
Step 4
Final Result
Multiply through.
$x^4 + 8x^3 + 24x^2 + 32x + 16$
Problem Statement
Find the 5th term in the expansion of $(3x - y^2)^7$.
General Term FormulaNegative Base Handling
Strategic Toolkit
$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$
Step 1
Assign Variables
Here $n=7$, $a=3x$, $b=-y^2$. For the 5th term, $r = 5 - 1 = 4$.
$T_5 = \binom{7}{4} (3x)^{3} (-y^2)^4$
Step 2
Evaluate Each Part
$\binom{7}{4} = 35$, $(3x)^3 = 27x^3$, $(-y^2)^4 = +y^8$ (even power, positive).
$35 \times 27x^3 \times y^8$
Step 3
Result
Multiply the numerical parts.
$T_5 = 945 x^3 y^8$
Problem Statement
Find the term independent of $x$ in the expansion of $(\frac{x}{2} + \frac{2}{x^2})^{12}$.
General TermIndependent TermExponent Equation
Strategic Toolkit
$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$
Step 1
Write General Term
With $a = \frac{x}{2}$ and $b = \frac{2}{x^2}$:
$T_{r+1} = \binom{12}{r} \left(\frac{x}{2}\right)^{12-r} \left(\frac{2}{x^2}\right)^r = \binom{12}{r} \frac{x^{12-r}}{2^{12-r}} \cdot \frac{2^r}{x^{2r}}$
Step 2
Simplify Powers of $x$
Combine the exponents of $x$.
$T_{r+1} = \binom{12}{r} \cdot 2^{2r-12} \cdot x^{12-3r}$
Step 3
Set Exponent to Zero
For the term independent of $x$: $12 - 3r = 0$.
$r = 4$
Step 4
Compute the Term
Substitute $r = 4$.
$T_5 = \binom{12}{4} \cdot 2^{-4} = 495 \cdot \frac{1}{16} = \frac{495}{16}$
Problem Statement
Expand $(1-2x)^{1/3}$ to four terms and use it to evaluate $(0.8)^{1/3}$ to 3 decimal places.
Binomial SeriesFractional IndexApproximation
Strategic Toolkit
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$$
Step 1
Identify Parameters
$(1-2x)^{1/3}$ has $n = 1/3$ and the substitution $x \to -2x$. Valid when $|2x| < 1$, i.e., $|x| < 1/2$.
$n = \frac{1}{3},\; \text{replace } x \text{ by } (-2x)$
Step 2
Expand to 4 Terms
Apply the binomial series formula carefully.
$1 + \frac{1}{3}(-2x) + \frac{\frac{1}{3}(-\frac{2}{3})}{2!}(-2x)^2 + \frac{\frac{1}{3}(-\frac{2}{3})(-\frac{5}{3})}{3!}(-2x)^3$
Step 3
Simplify
Evaluate each coefficient.
$= 1 - \frac{2}{3}x - \frac{4}{9}x^2 - \frac{40}{81}x^3 - \dots$
Step 4
Approximate $(0.8)^{1/3}$
Set $x = 0.1$ so that $1 - 2(0.1) = 0.8$.
$\approx 1 - 0.0667 - 0.0044 - 0.0005 = 0.928$
Problem Statement
Find the sum: $1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots$
Binomial SeriesSeries IdentificationSimultaneous Equations
Strategic Toolkit
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$$
Step 1
Compare with Binomial Series
Match the second term: $nx = \frac{1}{3}$. Match the third term: $\frac{n(n-1)}{2!}x^2 = \frac{1}{6}$.
$nx = \frac{1}{3}, \quad \frac{n(n-1)}{2}x^2 = \frac{1}{6}$
Step 2
Solve for $n$ and $x$
From the first equation, $x = \frac{1}{3n}$. Substituting into the second and solving gives $n = -\frac{1}{2}$.
$n = -\frac{1}{2}, \quad x = \frac{1}{3(-1/2)} = -\frac{2}{3}$
Step 3
Compute the Sum
The series is $(1 + x)^n = (1 - 2/3)^{-1/2} = (1/3)^{-1/2}$.
$= \sqrt{3}$
Memorize formulas
General Term
Gives any term directly. $r$ = term number minus 1.
Formula
$T_{r+1} = \binom{n}{r} a^{n-r} x^r$
Binomial Coefficient
The numerical weight of each term in the expansion.
Formula
$\binom{n}{r} = \frac{n!}{r!(n-r)!}$
Sum of All Coefficients
Set $a = 1, x = 1$ in $(a+x)^n$.
Formula
$\sum_{r=0}^{n} \binom{n}{r} = 2^n$
Binomial Series (Non-Integer Index)
Infinite expansion valid for $|x| < 1$.
Formula
$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$
Pascal's Identity
Each entry in Pascal's Triangle is the sum of two entries above.
Formula
$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$
Review
1
The term independent of $x$ in the expansion of $(\sqrt{x} + \frac{1}{2x^2})^{10}$ is:
Ans
$\frac{45}{4}$
Review
Tap to reveal
Set the power of x to zero
Write the general term and make the exponent of $x$ equal to zero, since the independent term has no $x$ factor.
$$T_{r+1}=\binom{10}{r}(\sqrt{x})^{10-r}\left(\frac{1}{2x^2}\right)^r=\binom{10}{r}\frac{1}{2^r}x^{(10-5r)/2}$$
Solve for the constant term
From $(10-5r)/2=0$, get $r=2$. Substituting gives the third term, whose coefficient is the required constant.
$$T_3=\binom{10}{2}\frac{1}{2^2}=45\cdot\frac{1}{4}=\frac{45}{4}$$
2
In the expansion of $(x^2 + \frac{1}{x})^9$, the term independent of $x$ is the $k$-th term. What is $k$?
Ans
$k = 7$ ($r = 6$, giving $T_7$)
Review
Tap to reveal
Form the general term
The $r$th choice contributes $x^{2(9-r)}$ from the first factor and $x^{-r}$ from the second, so the exponent of $x$ is $18-3r$.
$$T_{r+1}=\binom{9}{r}x^{18-3r}$$
Force the exponent to vanish
Set $18-3r=0$ to get $r=6$. Since the term number is $r+1$, the independent term is the seventh term.
$$k=r+1=7$$
3
The coefficient of $x^4$ in the expansion of $(3 - 2x)^7$ is:
Ans
$15120$
Review
Tap to reveal
Pick the term containing x^4
In the general term, the power of $x$ is exactly $r$, so the $x^4$ term comes from $r=4$.
$$T_{r+1}=\binom{7}{r}3^{7-r}(-2)^r x^r \Rightarrow T_5$$
Evaluate the coefficient
Substitute $r=4$ and note that the negative sign disappears because the power is even.
$$\binom{7}{4}3^3(-2)^4=35\cdot27\cdot16=15120$$
4
The sum $\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n}$ equals:
Ans
$n \cdot 2^{n-1}$
Review
Tap to reveal
Rewrite r\binom{n}{r}
Use the standard identity $r\binom{n}{r}=n\binom{n-1}{r-1}$ so every term shares the same factor $n$.
$$\sum_{r=1}^n r\binom{n}{r}=n\sum_{r=1}^n\binom{n-1}{r-1}$$
Apply the binomial sum
Reindex the sum and use $\sum_{j=0}^{n-1}\binom{n-1}{j}=2^{n-1}$.
$$n\sum_{j=0}^{n-1}\binom{n-1}{j}=n\cdot2^{n-1}$$
5
If the coefficient of $x^r$ in $(1+x)^{-2}$ is $(-1)^r(r+1)$, what is the coefficient of $x^5$?
Ans
$-6$
Review
Tap to reveal
Use the general coefficient
The coefficient of $x^r$ in $(1+x)^{-2}$ is given directly by the binomial-series pattern.
$$[x^r](1+x)^{-2}=(-1)^r(r+1)$$
Substitute r = 5
Plug in $r=5$ and evaluate the sign and magnitude together.
$$(-1)^5(5+1)=-6$$
6
The 3rd term from the end in the expansion of $(2x - \frac{1}{x})^9$ equals:
Ans
$-144x^{-5}$
Review
Tap to reveal
Convert from-end position to from-start position
For a binomial expansion with $n+1$ terms, the 3rd term from the end is the $(9-3+2)$th term from the beginning.
$$T_{9-3+2}=T_8$$
Evaluate the 8th term
Here $r=7$, so the term contains $(2x)^2$ and $(-1/x)^7$, giving a negative coefficient and power $x^{-5}$.
$$T_8=\binom{9}{7}(2x)^2\left(-\frac{1}{x}\right)^7=-144x^{-5}$$
7
The expansion of $(8 - 5x)^{-2/3}$ in ascending powers of $x$ is valid when:
Ans
$|x| < \frac{8}{5}$
Review
Tap to reveal
Factor into the standard binomial form
Pull out the constant 8 so the variable part becomes a power of $1+u$.
$$ (8-5x)^{-2/3}=8^{-2/3}\left(1-\frac{5x}{8}\right)^{-2/3} $$
Apply the convergence condition
The binomial series converges when the absolute value of the expansion variable is less than 1.
$$\left|\frac{5x}{8}\right|<1 \;\Rightarrow\; |x|<\frac{8}{5}$$
8
If $x$ is so small that $x^2$ and higher can be neglected, $\frac{1-x}{\sqrt{1+x}}$ is approximately:
Ans
$1 - \frac{3}{2}x$
Review
Tap to reveal
Expand the square-root denominator
Use the first-order binomial approximation $(1+x)^{-1/2}\approx1-\frac12x$ and keep only terms up to $x$.
$$\frac{1-x}{\sqrt{1+x}}=(1-x)(1+x)^{-1/2}\approx(1-x)\left(1-\frac12x\right)$$
Drop the x^2 term
Multiplying gives $1-\frac32x+\frac12x^2$, and the $x^2$ term is neglected by the problem statement.
$$1-\frac32x+\frac12x^2\approx1-\frac32x$$
9
The middle term(s) of $(\frac{3}{2}x - \frac{1}{3x})^{11}$ are at positions:
Ans
$T_6$ and $T_7$ (two middle terms since $n=11$ is odd)
Review
Tap to reveal
Count the total number of terms
A binomial expansion of $(\cdot)^{11}$ has $11+1=12$ terms, so there must be two middle terms.
$$\text{Total terms}=12$$
Find the two central positions
For an even number of terms, the middle terms are the 6th and 7th terms.
$$T_6\text{ and }T_7$$
10
The general term of the binomial series $(1+x)^{-3}$ for $|x| < 1$ is:
Ans
$(-1)^r \frac{(r+1)(r+2)}{2} x^r$
Review
Tap to reveal
Write the general binomial term
For $(1+x)^{-3}$, the $r$th term is obtained from the falling product $(-3)(-4)\cdots(-3-r+1)$ divided by $r!$.
$$T_{r+1}=\frac{(-3)(-4)\cdots(-3-r+1)}{r!}x^r$$
Simplify the coefficient
Pull out $(-1)^r$ and rewrite the positive product as a factorial ratio to get the closed form.
$$T_{r+1}=(-1)^r\frac{(r+1)(r+2)}{2}x^r$$
11
In the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$, the second and third terms equal $\frac{1}{3}$ and $\frac{1}{9}$ respectively. The value of $n$ is:
Ans
$-1$
Review
Tap to reveal
Use the second term to express x
The coefficient of $x$ is $n$, so matching it to $\frac13$ gives $x=\frac{1}{3n}$.
$$nx=\frac13\;\Rightarrow\;x=\frac{1}{3n}$$
Substitute into the third term
Put that value into $\frac{n(n-1)}{2}x^2=\frac19$ and solve the resulting linear equation in $n$.
$$\frac{n(n-1)}{2}\left(\frac{1}{3n}\right)^2=\frac19\;\Rightarrow\;n=-1$$
12
In the expansion of $(1+x)^{2n}$, the middle term can be written as $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \cdot 2^n \cdot x^n$. This uses the identity:
Ans
$\binom{2n}{n} = \frac{(2n)!}{(n!)^2} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} \cdot 2^n$
Review
Tap to reveal
Identify the middle term
The middle term of $(1+x)^{2n}$ is the $(n+1)$th term, so its coefficient is $\binom{2n}{n}$.
$$T_{n+1}=\binom{2n}{n}x^n$$
Rewrite the central binomial coefficient
Split $(2n)!$ into odd and even factors and use $2\cdot4\cdots2n=2^n n!$.
$$\binom{2n}{n}=\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\cdot2^n$$
Question 1 of 12
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1
The term independent of $x$ in the expansion of $(\sqrt{x} + \frac{1}{2x^2})^{10}$ is: