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Mathematics AA · HL · Unit 1: Number and Algebra

Unit 1Topic 1.9Key Concepts35 min read

Binomial theorem

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Introduction to the binomial theorem

A binomial is an expression with two terms, such as x+3x+3, 2a−b2a-b, or 1+x21+\frac{x}{2}. The binomial theorem expands a non-negative integer power (a+b)n(a+b)^n without repeated multiplication.

For a nonnegative integer exponent, the binomial formula contains n+1n+1 formal summands before collecting like terms or substituting values; fewer nonzero terms may remain. The power of the first component starts at nn and decreases by 1, while the power of the second component starts at 0 and increases by 1. In every term, the two exponents add to nn.

The numerical multipliers are binomial coefficients. They can be read from Pascal’s triangle or calculated using (nr)\binom nr.

At AA SL, the finite binomial theorem uses n∈{0,1,2,…}n\in\{0,1,2,\ldots\}. A negative or fractional number may appear inside the binomial as a coefficient, but a negative or fractional exponent requires a different generalized theorem.

Key terms

Binomial
An algebraic expression consisting of two terms.
Expansion
An equivalent sum obtained by removing brackets and collecting terms.
Binomial coefficient
A number (nr)\binom nr multiplying a term in the expansion.
Non-negative integer
One of 0,1,2,3,…0,1,2,3,\ldots; the allowed powers in the finite AA SL theorem.

Worked examples

Recognize the structure

How many terms are in the expansion of (x+2)7(x+2)^7?

  1. 1For power nn, a binomial expansion contains n+1n+1 terms.
  2. 2Here n=7n=7.

Answer: There are 88 terms.

Track the exponents

Describe the powers of xx and yy in (x+y)4(x+y)^4.

  1. 1The power of xx decreases from 4 to 0.
  2. 2The power of yy increases from 0 to 4.
  3. 3Each pair of powers adds to 4.

Answer: The power pairs are (4,0),(3,1),(2,2),(1,3),(0,4)(4,0),(3,1),(2,2),(1,3),(0,4).

Expand by known pattern

Expand (a+b)2(a+b)^2.

  1. 1The coefficients are 1,2,11,2,1.
  2. 2Match them with a2,a1b1,b2a^2,a^1b^1,b^2.

Answer: (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2.

Rules and formulas

Binomial theorem

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r
nn
a non-negative integer power
rr
the term index running from 0 to n
a,ba,b
the two components of the binomial

Number of formal binomial summands

formal summands before collection=n+1\text{formal summands before collection}=n+1
nn
nonnegative integer exponent; like terms may combine and some summands may vanish

Pascal’s triangle and coefficient patterns

Pascal’s triangle organizes the coefficients of (a+b)n(a+b)^n. Row n=0n=0 is 11. Each new row begins and ends with 1, and every interior entry is the sum of the two entries above it.

The first rows are n=0:1n=0: 1; n=1:1,1n=1: 1,1; n=2:1,2,1n=2: 1,2,1; n=3:1,3,3,1n=3: 1,3,3,1; n=4:1,4,6,4,1n=4: 1,4,6,4,1; n=5:1,5,10,10,5,1n=5: 1,5,10,10,5,1; and n=6:1,6,15,20,15,6,1n=6: 1,6,15,20,15,6,1.

Every row is symmetric because (nr)=(nn−r)\binom nr=\binom n{n-r}. The sum of all coefficients in row nn is 2n2^n, found by substituting a=b=1a=b=1 into (a+b)n(a+b)^n.

For n≥1n\ge1, the alternating sum of a row is zero. This follows by substituting a=1,b=−1a=1,b=-1, giving (1−1)n=0(1-1)^n=0.

Key terms

Pascal’s triangle
A triangular arrangement whose row nn contains the coefficients of (a+b)n(a+b)^n.
Row index
The value nn, beginning with the single 1 as row 0.
Symmetry
The matching coefficient pattern from left and right.
Adjacent-sum rule
Each interior entry equals the sum of the two entries directly above it.

Worked examples

Build the next row

Use row 1,5,10,10,5,11,5,10,10,5,1 to form the next Pascal row.

  1. 1Place 1 at each end.
  2. 2Add adjacent entries: 1+5=61+5=6, 5+10=155+10=15, 10+10=2010+10=20, then use symmetry.

Answer: 1,6,15,20,15,6,11,6,15,20,15,6,1.

Use a Pascal row

Write the coefficient pattern for (p+q)5(p+q)^5.

  1. 1Use row n=5n=5 of Pascal’s triangle.

Answer: The coefficients are 1,5,10,10,5,11,5,10,10,5,1.

Check a coefficient sum

Find the sum of the coefficients in (x+y)8(x+y)^8.

  1. 1Substitute x=1,y=1x=1,y=1.
  2. 2The expansion becomes (1+1)8(1+1)^8.

Answer: The coefficient sum is 28=2562^8=256.

Rules and formulas

Pascal recurrence

(nr)=(n−1r−1)+(n−1r)\binom nr=\binom{n-1}{r-1}+\binom{n-1}{r}

Coefficient symmetry

(nr)=(nn−r)\binom nr=\binom n{n-r}

Row sum

∑r=0n(nr)=2n\sum\limits_{r=0}^{n}\binom nr=2^n

Combination notation and binomial coefficients

The symbol (nr)\binom nr, read “nn choose rr,” counts the ways to choose rr objects from nn when order does not matter. In the binomial theorem it is the coefficient of an−rbra^{n-r}b^r.

For integers 0≤r≤n0\le r\le n, (nr)=n!r!(n−r)!\binom nr=\frac{n!}{r!(n-r)!}. The factorial n!n! means n(n−1)(n−2)⋯2⋅1n(n-1)(n-2)\cdots2\cdot1, with 0!=10!=1.

A GDC usually provides an nCrnCr command. Enter nn, choose nCrnCr, then enter rr. Even when using technology, show which coefficient is being calculated.

Key terms

Combination
A selection in which order does not matter.
Factorial
n!=n(n−1)⋯1n!=n(n-1)\cdots1, with 0!=10!=1.
nCr
The calculator command for (nr)\binom nr.
Boundary coefficient
(n0)=(nn)=1\binom n0=\binom nn=1, the first and last coefficients.

Worked examples

Calculate a coefficient

Evaluate (72)\binom72.

  1. 1Use (72)=7![2!5!]\binom72=\frac{7!}{[2!5!]}.
  2. 2Cancel 5!5!: 7⋅6(2⋅1)\frac{7\cdot6}{(2\cdot1)}.

Answer: (72)=21\binom72=21.

Use symmetry

Evaluate (108)\binom{10}{8} efficiently.

  1. 1Use symmetry: (108)=(102)\binom{10}{8}=\binom{10}{2}.
  2. 2Calculate 10⋅92\frac{10\cdot9}{2}.

Answer: 4545.

GDC nCr method

Find (156)\binom{15}{6} with a GDC.

  1. 1Enter 1515.
  2. 2Select the combination command nCrnCr.
  3. 3Enter 66 and evaluate.

Answer: (156)=5005\binom{15}{6}=5005.

Rules and formulas

Combination formula

(nr)=n!r!(n−r)!\binom nr=\frac{n!}{r!(n-r)!}

Factorial

n!=n(n−1)(n−2)⋯2⋅1,0!=1n!=n(n-1)(n-2)\cdots2\cdot1,\qquad 0!=1

Boundary values

(n0)=(nn)=1\binom n0=\binom nn=1

Expanding (a+b)n\left(a + b\right)^{n}

To expand a binomial, identify the complete first and second components, choose the coefficients for row nn, decrease the first power, increase the second power, then simplify each term.

Keep a negative component in parentheses as (−k)r(-k)^r, with k>0k>0. This factor is negative for odd rr and positive for even rr. The sign of the whole term also depends on its other factors; for (a−kx)n(a-kx)^n with a,k>0a,k>0, the coefficients of successive powers of xx alternate.

Arrange the final polynomial in descending powers of the chosen variable unless the question requests another order.

Key terms

Component
One of the two complete terms inside the binomial brackets.
Descending powers
Polynomial order from highest exponent to lowest exponent.
Alternating signs
A sign pattern caused by powers of a negative binomial component.

Worked examples

Basic expansion

Expand (x+2)4(x+2)^4.

  1. 1Use coefficients 1,4,6,4,11,4,6,4,1.
  2. 2Form x4+4x3(2)+6x2(22)+4x(23)+24x^4+4x^3(2)+6x^2(2^2)+4x(2^3)+2^4.
  3. 3Simplify each coefficient.

Answer: x4+8x3+24x2+32x+16x^4+8x^3+24x^2+32x+16.

Negative second component

Expand (2x−3)3(2x-3)^3.

  1. 1Use coefficients 1,3,3,11,3,3,1.
  2. 2Form (2x)3+3(2x)2(−3)+3(2x)(−3)2+(−3)3(2x)^3+3(2x)^2(-3)+3(2x)(-3)^2+(-3)^3.
  3. 3Simplify while preserving signs.

Answer: 8x3−36x2+54x−278x^3-36x^2+54x-27.

Reverse-looking binomial

Expand (3−x)4(3-x)^4 in descending powers of xx.

  1. 1Use components a=3a=3, b=−xb=-x.
  2. 2The raw order begins with 343^4 and ends with (−x)4(-x)^4.
  3. 3Simplify, then reorder by powers of xx.

Answer: x4−12x3+54x2−108x+81x^4-12x^3+54x^2-108x+81.

Two variable components

Expand (a+2b)3(a+2b)^3.

  1. 1Use coefficients 1,3,3,11,3,3,1.
  2. 2Form a3+3a2(2b)+3a(2b)2+(2b)3a^3+3a^2(2b)+3a(2b)^2+(2b)^3.

Answer: a3+6a2b+12ab2+8b3a^3+6a^2b+12ab^2+8b^3.

Rules and formulas

Expanded form

(a+b)n=(n0)an+(n1)an−1b+⋯+(nn)bn(a+b)^n=\binom n0a^n+\binom n1a^{n-1}b+\cdots+\binom nnb^n

Exponent check

(n−r)+r=n(n-r)+r=n

The general term of a binomial expansion

The term containing brb^r is Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r. The term number is r+1r+1 because the first term corresponds to r=0r=0.

To find the ppth term, set r=p−1r=p-1. This off-by-one adjustment is one of the most important details in binomial questions.

The general term lets us find one required term without writing the full expansion. It also exposes the power of a variable, which can be matched to a requested exponent.

Key terms

General term
A formula describing every term of the expansion using an index rr.
Term number
The position r+1r+1 of a term when rr begins at zero.
Index
The integer rr ranging from 0 to nn.

Worked examples

Write a general term

Write the general term of (2x+3)7(2x+3)^7.

  1. 1Use a=2xa=2x, b=3b=3, and n=7n=7.
  2. 2Substitute into Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r.

Answer: Tr+1=(7r)(2x)7−r3rT_{r+1}=\binom7r(2x)^{7-r}3^r.

Find the fourth term

Find the fourth term of (2x−y)6(2x-y)^6.

  1. 1For the fourth term, r=4−1=3r=4-1=3.
  2. 2T4=(63)(2x)3(−y)3T_4=\binom63(2x)^3(-y)^3.
  3. 3(63)=20\binom63=20, (2x)3=8x3(2x)^3=8x^3, and (−y)3=−y3(-y)^3=-y^3.

Answer: T4=−160x3y3T_4=-160x^3y^3.

Find the final term

Find the eighth term of (a+2b)7(a+2b)^7.

  1. 1There are 7+1=87+1=8 terms, so the eighth uses r=7r=7.
  2. 2T8=(77)a0(2b)7T_8=\binom77a^0(2b)^7.

Answer: T8=128b7T_8=128b^7.

Check an index range

What values of rr occur in (p+q)5(p+q)^5?

  1. 1The index begins at 0 and ends at n=5n=5.

Answer: r=0,1,2,3,4,5r=0,1,2,3,4,5, producing six terms.

Rules and formulas

General term

Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

pth-term index

r=p−1r=p-1

Finding a specific term or coefficient

To find the coefficient of xkx^k, write the general term, determine its power of xx, set that exponent equal to kk, and solve for the integer index rr. Then evaluate only that term.

The coefficient does not include the variable factor. After finding the matching term, remove xkx^k and report its numerical or parameter coefficient.

If the equation for rr gives a non-integer or a value outside 0≤r≤n0\le r\le n, no such term appears.

Key terms

Coefficient of xᵏ
The factor multiplying xkx^k after like terms are collected.
Power matching
Equating the variable exponent in the general term to the requested exponent.
Admissible index
An integer rr satisfying 0≤r≤n0\le r\le n.

Worked examples

Coefficient of x cubed

Find the coefficient of x3x^3 in (2+x)7(2+x)^7.

  1. 1The general term is (7r)27−rxr\binom7r2^{7-r}x^r.
  2. 2For x3x^3, set r=3r=3.
  3. 3The coefficient is (73)24=35(16)\binom73 2^4=35(16).

Answer: The coefficient is 560560.

Power from the first component

Find the coefficient of x4x^4 in (3x−2)6(3x-2)^6.

  1. 1Tr+1=(6r)(3x)6−r(−2)rT_{r+1}=\binom6r(3x)^{6-r}(-2)^r.
  2. 2The power of xx is 6−r6-r, so 6−r=46-r=4 gives r=2r=2.
  3. 3Coefficient =(62)34(−2)2=15(81)(4)=\binom62 3^4(-2)^2=15(81)(4).

Answer: The coefficient is 48604860.

Find a parameter

The coefficient of xx in (1+kx)6(1+kx)^6 is 18. Find kk.

  1. 1The xx term uses r=1r=1.
  2. 2Its coefficient is (61)k=6k\binom61k=6k.
  3. 3Solve 6k=186k=18.

Answer: k=3k=3.

Find the exponent n

The coefficient of x2x^2 in (1+x)n(1+x)^n is 45. Find the positive integer nn.

  1. 1The coefficient is (n2)=n(n−1)2\binom n2=\frac{n(n-1)}{2}.
  2. 2Solve n(n−1)2=45\frac{n(n-1)}{2}=45, giving n2−n−90=0n^2-n-90=0.
  3. 3Factor: (n−10)(n+9)=0(n-10)(n+9)=0.
  4. 4Reject n=−9n=-9 because nn is a positive integer.

Answer: n=10n=10.

Rules and formulas

Coefficient method

power of x=k⟹solve for r\text{power of }x=k\quad\Longrightarrow\quad \text{solve for }r

Quadratic coefficient

(n2)=n(n−1)2\binom n2=\frac{n(n-1)}2

Finding constant terms

A constant term has variable exponent zero. In an expansion involving positive and negative powers of xx, write the general term and collect the total exponent of xx.

Set that exponent equal to zero and solve for rr. A constant exists only if the solution is an integer in 0≤r≤n0\le r\le n.

After finding rr, substitute it into the entire general term. Do not stop after solving the exponent equation; the constant coefficient still has to be evaluated.

Key terms

Constant term
A term independent of the variable, equivalently containing x0=1x^0=1.
Net exponent
The total exponent obtained after combining all powers of the variable in a term.
Integer-index test
The requirement that a binomial term index be a whole number between 0 and nn.

Worked examples

Constant from mixed powers

Find the constant term in (2x2+1x)6(2x^2+\frac{1}{x})^6.

  1. 1Tr+1=(6r)(2x2)6−r(x−1)rT_{r+1}=\binom6r(2x^2)^{6-r}(x^{-1})^r.
  2. 2The exponent of xx is 2(6−r)−r=12−3r2(6-r)-r=12-3r.
  3. 3Set 12−3r=012-3r=0, giving r=4r=4.
  4. 4The constant is (64)26−4=15(4)\binom64 2^{6-4}=15(4).

Answer: The constant term is 6060.

Another constant-term equation

Find the constant term in (x3+2x2)5(x^3+\frac{2}{x^2})^5.

  1. 1The general term has xx-exponent 3(5−r)−2r=15−5r3(5-r)-2r=15-5r.
  2. 2Set 15−5r=015-5r=0, so r=3r=3.
  3. 3The constant coefficient is (53)23=10(8)\binom53 2^3=10(8).

Answer: The constant term is 8080.

Show that no constant exists

Does (x+x−2)4(x+x^{-2})^4 contain a constant term?

  1. 1The net exponent is (4−r)−2r=4−3r(4-r)-2r=4-3r.
  2. 2Setting it to zero gives r=43r=\frac{4}{3}.
  3. 3This is not an integer index.

Answer: No constant term occurs.

Constant from an ordinary polynomial

Find the constant term in (3x−2)7(3x-2)^7.

  1. 1The only constant occurs when the power of 3x3x is zero, so 6−r=06-r=0 would be wrong because n=7n=7; use 7−r=07-r=0.
  2. 2Thus r=7r=7, giving (77)(3x)0(−2)7\binom77(3x)^0(-2)^7.

Answer: The constant term is −128-128.

Rules and formulas

Constant-term condition

net exponent of x=0\text{net exponent of }x=0

Admissible index

r∈{0,1,2,…,n}r\in\{0,1,2,\ldots,n\}

Negative or fractional coefficients and GDC methods

Negative and fractional components follow the same finite theorem. Keep a negative component in parentheses so its powers produce the correct sign, and keep fractional powers of coefficients exact.

For (a−kx)n(a-kx)^n with a,k>0a,k>0 and nonnegative integer nn, coefficients alternate in sign with rr, because of (−k)r(-k)^r. For (a+xm)n(a+\frac{x}{m})^n, with m≠0m\ne0, the factor (xm)r(\frac{x}{m})^r contributes m−rm^{-r} to each coefficient.

A GDC can calculate nCrnCr, factorials, or individual terms. A CAS expansion may be useful as a check where permitted, but an IB response should show the theorem or general-term method.

Write P(x)=c0+c1x+⋯+cnxnP(x)=c_0+c_1x+\cdots+c_nx^n, including zero coefficients for missing powers. Then P(1)=c0+c1+⋯+cnP(1)=c_0+c_1+\cdots+c_n, while P(−1)=c0−c1+c2−⋯+(−1)ncnP(-1)=c_0-c_1+c_2-\cdots+(-1)^n c_n; signs follow exponent parity, not displayed-term order.

Key terms

Fractional coefficient
A component such as x2\frac{x}{2}, whose powers contribute fractional numerical factors.
Coefficient sum
The sum of all polynomial coefficients, found by evaluating the polynomial at x=1x=1.
Alternating coefficient sum
For P(x)=∑k=0nckxkP(x)=\sum\limits_{k=0}^n c_kx^k, the alternating coefficient sum is ∑k=0n(−1)kck=P(−1)\sum\limits_{k=0}^n(-1)^k c_k=P(-1). Include zero coefficients for missing powers and start with the positive constant coefficient.
CAS
A computer algebra system that can manipulate symbolic expressions where its use is allowed.

Worked examples

Fractional component expansion

Expand (x+12)4(x+\frac{1}{2})^4.

  1. 1Use coefficients 1,4,6,4,11,4,6,4,1.
  2. 2Form x4+4x3(12)+6x2(12)2+4x(12)3+(12)4x^4+4x^3(\frac{1}{2})+6x^2(\frac{1}{2})^2+4x(\frac{1}{2})^3+(\frac{1}{2})^4.

Answer: x4+2x3+32x2+12x+116x^4+2x^3+\frac32x^2+\frac12x+\frac1{16}.

Alternating expansion

Expand (1−2x)5(1-2x)^5.

  1. 1Use row 1,5,10,10,5,11,5,10,10,5,1.
  2. 2Evaluate powers of (−2x)r(-2x)^r, preserving the alternating signs.

Answer: 1−10x+40x2−80x3+80x4−32x51-10x+40x^2-80x^3+80x^4-32x^5.

Sum of coefficients

Find the sum of the coefficients in (2x−3)5(2x-3)^5 without expanding.

  1. 1Set x=1x=1.
  2. 2The polynomial value is (2−3)5=(−1)5(2-3)^5=(-1)^5.

Answer: The coefficient sum is −1-1.

GDC coefficient calculation

Find the coefficient of x7x^7 in (1+3x)12(1+3x)^{12}.

  1. 1The matching index is r=7r=7.
  2. 2Coefficient =(127)37=\binom{12}{7}3^7.
  3. 3Use 12 nCr 712\ nCr\ 7 on the GDC, then multiply by 373^7.

Answer: The coefficient is 1 732 1041\,732\,104.

Rules and formulas

Sum of coefficients

∑k=0nck=P(1)\sum\limits_{k=0}^{n}c_k=P(1)

Alternating coefficient sum

∑k=0n(−1)kck=P(−1)\sum\limits_{k=0}^{n}(-1)^kc_k=P(-1)

GDC term calculation

(nr)an−rbr\binom nr a^{n-r}b^r

IB-style practice with worked solutions

Attempt each question before opening its solution. The problems progress from Pascal coefficients and full expansions to targeted terms, constants, parameters, and coefficient identities.

For targeted questions, avoid expanding everything. Write the general term, match the required power, verify the index, then simplify only the necessary term.

Key terms

Targeted expansion
Finding one requested term or coefficient without writing every term.
Parameter
An unknown constant determined from a coefficient condition.
Verification
Checking end terms, exponent sums, signs, symmetry, or substitution values.

Worked examples

Practice 1 — full expansion

Expand (x+3)3(x+3)^3.

  1. 1Use coefficients 1,3,3,11,3,3,1.
  2. 2Form x3+3x2(3)+3x(32)+33x^3+3x^2(3)+3x(3^2)+3^3.

Answer: x3+9x2+27x+27x^3+9x^2+27x+27.

Practice 2 — Pascal row

State the coefficients of (a+b)6(a+b)^6.

  1. 1Use Pascal row n=6n=6.

Answer: 1,6,15,20,15,6,11,6,15,20,15,6,1.

Practice 3 — combination

Evaluate (92)\binom92.

  1. 1(92)=9⋅82\binom92=\frac{9\cdot8}{2}.

Answer: 3636.

Practice 4 — specified term

Find the fifth term of (a+2b)7(a+2b)^7.

  1. 1The fifth term uses r=4r=4.
  2. 2T5=(74)a3(2b)4=35a3(16b4)T_5=\binom74a^3(2b)^4=35a^3(16b^4).

Answer: 560a3b4560a^3b^4.

Practice 5 — coefficient of x cubed

Find the coefficient of x3x^3 in (1+4x)5(1+4x)^5.

  1. 1Use r=3r=3.
  2. 2Coefficient =(53)43=10(64)=\binom534^3=10(64).

Answer: 640640.

Practice 6 — negative component

Find the coefficient of x2x^2 in (3−2x)6(3-2x)^6.

  1. 1Use r=2r=2.
  2. 2Coefficient =(62)34(−2)2=15(81)(4)=\binom623^{4}(-2)^2=15(81)(4).

Answer: 48604860.

Practice 7 — constant term

Find the constant term in (x2+3x)6(x^2+\frac{3}{x})^6.

  1. 1The net exponent is 2(6−r)−r=12−3r2(6-r)-r=12-3r.
  2. 2Set 12−3r=012-3r=0, so r=4r=4.
  3. 3Constant =(64)34=15(81)=\binom643^4=15(81).

Answer: 12151215.

Practice 8 — no constant term

Does (x+x−2)4(x+x^{-2})^4 contain a constant?

  1. 1The net exponent is 4−3r4-3r.
  2. 2Solving 4−3r=04-3r=0 gives r=43r=\frac{4}{3}, not an integer.

Answer: No constant term occurs.

Practice 9 — fractional coefficient

Expand (1−x2)4(1-\frac{x}{2})^4.

  1. 1Use coefficients 1,4,6,4,11,4,6,4,1.
  2. 2Evaluate powers of −x2\frac{-x}{2}.

Answer: 1−2x+32x2−12x3+116x41-2x+\frac32x^2-\frac12x^3+\frac1{16}x^4.

Practice 10 — find n

The coefficient of x2x^2 in (1+x)n(1+x)^n is 45. Find nn.

  1. 1Set (n2)=45\binom n2=45.
  2. 2Solve n(n−1)=90n(n-1)=90.
  3. 3The positive integer solution is n=10n=10.

Answer: n=10n=10.

Practice 11 — coefficient sum

Find the sum of coefficients in (2x−3)5(2x-3)^5.

  1. 1Set x=1x=1.
  2. 2Evaluate (2−3)5(2-3)^5.

Answer: −1-1.

Practice 12 — alternating coefficient sum

Find the alternating sum of coefficients in (1+x)8(1+x)^8.

  1. 1Set x=−1x=-1.
  2. 2Evaluate (1−1)8(1-1)^8.

Answer: 00.

Practice 13 — parameter

The coefficient of xx in (1+kx)6(1+kx)^6 is 18. Find kk.

  1. 1The xx coefficient is (61)k=6k\binom61k=6k.
  2. 2Solve 6k=186k=18.

Answer: k=3k=3.

Rules and formulas

Core theorem

(a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r

Target term

Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r

Constant condition

net power of x=0\text{net power of }x=0

Topic summary

  • For non-negative integer nn, (a+b)n=∑r=0n(nr)an−rbr(a+b)^n=\sum\limits_{r=0}^{n}\binom nr a^{n-r}b^r.
  • For a nonnegative integer nn, the binomial formula has n+1n+1 formal summands before like-term collection; the simplified result may have fewer nonzero terms.
  • Pascal row nn supplies the coefficients of (a+b)n(a+b)^n.
  • (nr)=n![r!(n−r)!]\binom nr=\frac{n!}{[r!(n-r)!]} and (nr)=(nn−r)\binom nr=\binom n{n-r}.
  • The general term is Tr+1=(nr)an−rbrT_{r+1}=\binom nr a^{n-r}b^r.
  • For the ppth term, use r=p−1r=p-1.
  • To find the coefficient of xkx^k, match the general-term exponent to kk.
  • A constant term requires a net variable exponent of zero and an integer rr in [0,n][0,n].
  • Keep negative binomial components in parentheses so odd and even powers produce correct signs.
  • Fractional components are allowed; keep their coefficients exact.
  • Write P(x)=c0+c1x+⋯+cnxnP(x)=c_0+c_1x+\cdots+c_nx^n, including zero coefficients for missing powers. Then P(1)=c0+c1+⋯+cnP(1)=c_0+c_1+\cdots+c_n, while P(−1)=c0−c1+c2−⋯+(−1)ncnP(-1)=c_0-c_1+c_2-\cdots+(-1)^n c_n; signs follow exponent parity, not displayed-term order.
  • A GDC can evaluate nCrnCr, but an IB solution should still show the selected term and method.