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Equations · Algebra

Newton's Binomial Theorem

Before you start, make sure you know: Properties of exponents (High School Year 1 (semesters 1 and 2))
Before you start, make sure you know: Combinations (High School Year 1 (semesters 1 and 2))

Expanding $(2x-y)^4$ by multiplying parentheses by hand is long, and it's easy to slip up on the signs. The binomial theorem gives you the coefficients of that expansion without multiplying anything: just by counting combinations. The same idea counts the paths of a marble on a Galton board or the genotypes in a population, and it shows up whenever something is repeated $n$ times with two possible outcomes.

Factorials and binomial coefficients

The factorial of a natural number, $n!$, is the product of all the natural numbers from $1$ up to $n$ ($0!=1$ by convention). The binomial coefficient $\binom{n}{k}$ (in Mexico, also "combinations of $n$ taken $k$ at a time") counts the number of ways to choose $k$ elements from a total of $n$, regardless of order.

Detailed calculation of 7 factorial and of the binomial coefficient 7 choose 3. 7! = 7·6·5·4·3·2·1 = 5,040 C(7,3) = 7! / (3!·4!) = 5,040 / (6·24) = 35
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Calculate the value of the following expression using the properties of factorials and binomial coefficients: $\frac{10!}{8! \times 2!} + \binom{7}{3}$.

Calculate the value of the following expression using the properties of factorials and binomial coefficients: $\frac{10!}{8! \times 2!} + \binom{7}{3}$.

Solve

Pascal's triangle

The numbers $\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}$ in row $n$ are the coefficients of the expansion of $(a+b)^n$. Arranged in a triangle, each number is the sum of the two above it: this is Pascal's triangle (also called Tartaglia's triangle).

Pascal's triangle with rows n=0 to n=6; row n=4 is highlighted. n=01 n=111 n=2121 n=31331 n=4 14641 n=515101051 n=61615201561 row n=4: coefficients of (a+b)⁴
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Use Pascal's triangle (or Tartaglia's triangle) to obtain the coefficients of the sixth row and expand the expression $(m + 2n)^6$.

Use Pascal's triangle (or Tartaglia's triangle) to obtain the coefficients of the sixth row and expand the expression $(m + 2n)^6$.

Solve

Equations with factorials and binomial coefficients

When the top or bottom number is an unknown, you have to expand the factorials until they cancel and end up with an ordinary equation (almost always linear or quadratic).

Strategy: expand the larger factorial down to the smaller one, cancel, and solve the equation. (x+1)! = (x+1)·x·(x−1)! C(x,2) = x·(x−1) / 2 Expand, cancel, solve, and check that x is natural
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Simplify as far as possible the following algebraic expression involving factorials: $\frac{(n+2)!}{n! \times (n+1)}$.

Simplify as far as possible the following algebraic expression involving factorials: $\frac{(n+2)!}{n! \times (n+1)}$.

Solve
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Find the value of the natural number $x$ satisfying the following equation involving binomial coefficients: $\binom{x}{2} = 21$.

Find the value of the natural number $x$ satisfying the following equation involving binomial coefficients: $\binom{x}{2} = 21$.

Solve
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Solve the following equation involving binomial coefficients, finding the value of the natural number $x$: $\binom{x}{3} + \binom{x}{2} = 4 \binom{x}{1}$.

Solve the following equation involving binomial coefficients, finding the value of the natural number $x$: $\binom{x}{3} + \binom{x}{2} = 4 \binom{x}{1}$.

Solve
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Prove the fundamental property of binomial coefficients $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ using the algebraic definition with factorials. Subsequently,…

Prove the fundamental property of binomial coefficients $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ using the algebraic definition with factorials. Subsequently, use the properties of binomial coefficients to solve the following system of equations for natural numbers $x$ and $y$:

$\binom{x}{y} = \binom{x}{y+1}$

$4x - 5y = 19$

Solve
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Solve the following algebraic equation involving binomial coefficients and factorials, finding the value of the natural number $x$:…

Solve the following algebraic equation involving binomial coefficients and factorials, finding the value of the natural number $x$: $\frac{(x+1)!}{(x-1)!} - 2\binom{x}{2} = 14 + \binom{x-1}{1}$.

Solve

The binomial theorem formula

Each term of the expansion of $(a+b)^n$ combines a decreasing power of $a$, an increasing power of $b$, and the corresponding coefficient from Pascal's triangle.

Outline of the expansion of (x+3)⁴ with coefficients from the triangle and ordered powers. (x+3)⁴: coefficients 1, 4, 6, 4, 1 powers of x: 4,3,2,1,0 — powers of 3: 0,1,2,3,4 x⁴ + 12x³ + 54x² + 108x + 81
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Fully expand the binomial $(x + 3)^4$ using the binomial theorem.

Fully expand the binomial $(x + 3)^4$ using the binomial theorem.

Solve
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Expand the expression $(2a - b)^5$ and state how many terms the final expansion has.

Expand the expression $(2a - b)^5$ and state how many terms the final expansion has.

Solve
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Expand the binomial with fractions $\left(x + \frac{1}{2}\right)^3$ and simplify each of the resulting terms.

Expand the binomial with fractions $\left(x + \frac{1}{2}\right)^3$ and simplify each of the resulting terms.

Solve

The general term

If you're only asked for "the fourth term" or "the term in $x^3$," you don't need to expand everything: the general term formula gives it directly.

Calculation of the fourth term of (x² − 2y)⁷ using the general term. (x² − 2y)⁷ fourth term: position 4, so k = 3 T₄ = C(7,3) (x²)⁴ (−2y)³ = 35 · x⁸ · (−8y³) = −280x⁸y³
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Write down directly the fourth term in the expansion of the binomial $(x^2 - 2y)^7$, without finding the full expansion.

Write down directly the fourth term in the expansion of the binomial $(x^2 - 2y)^7$, without finding the full expansion.

Solve
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Find the coefficient of the term containing $x^3$ in the expansion of the binomial $(x - 4)^5$.

Find the coefficient of the term containing $x^3$ in the expansion of the binomial $(x - 4)^5$.

Solve
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Find the term independent of $x$ (that is, the term that does not contain the variable $x$) in the expansion of the binomial $\left(x^2 - \frac{1}{x}\right)^9$.

Find the term independent of $x$ (that is, the term that does not contain the variable $x$) in the expansion of the binomial $\left(x^2 - \frac{1}{x}\right)^9$.

Solve

Radicals, fractions, and the sought term

The general term works just as well when $a$ or $b$ are fractions, roots, or carry their own sign. You have to reduce the exponent of $x$ correctly using the properties of exponents, and then set it equal to the exponent you're looking for (for the constant term, to $0$).

How to find the constant term of (x² − 1/x)⁹ using the exponent equation. (x² − 1/x)⁹: T(k+1) = C(9,k)(x²)^(9−k)(−1/x)^k exponent of x: 2(9−k) − k = 18 − 3k constant term: 18 − 3k = 0 → k = 6
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Determine the middle term in the expansion of the binomial $\left(\sqrt{x} - \frac{2}{x}\right)^6$ and simplify the resulting expression.

Determine the middle term in the expansion of the binomial $\left(\sqrt{x} - \frac{2}{x}\right)^6$ and simplify the resulting expression.

Solve
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Find the fifth term in the expansion of the binomial $\left(2x^2 - \frac{1}{x}\right)^7$. Simplify the result by combining powers of $x$.

Find the fifth term in the expansion of the binomial $\left(2x^2 - \frac{1}{x}\right)^7$. Simplify the result by combining powers of $x$.

Solve
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Expand and fully simplify the expression involving radicals $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^4$. Collect like terms and leave the result without negative…

Expand and fully simplify the expression involving radicals $\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^4$. Collect like terms and leave the result without negative exponents.

Solve

Finding a parameter

Sometimes the binomial contains an unknown letter and you are given information about one term: its coefficient or its value. You write the general term, set it equal to the given data, and solve the resulting equation.

Steps to find a parameter: general term, set equal to the data, and solve. 1. General term with the parameter 2. Set it equal to the data (coefficient or term) 3. Solve the equation and check the conditions
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In the expansion of the binomial $(x + k)^4$, the coefficient of the term in $x^2$ is known to be equal to $54$. Calculate the positive value of the parameter $k$.

In the expansion of the binomial $(x + k)^4$, the coefficient of the term in $x^2$ is known to be equal to $54$. Calculate the positive value of the parameter $k$.

Solve
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Find the value of the real parameter $k$ such that the coefficient of the term in $x^3$ in the expansion of the binomial $\left(x^2 + \frac{k}{x}\right)^6$ is exactly…

Find the value of the real parameter $k$ such that the coefficient of the term in $x^3$ in the expansion of the binomial $\left(x^2 + \frac{k}{x}\right)^6$ is exactly equal to $-160$.

Solve
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Determine the positive real values of the parameters $a$ and $b$ in the binomial $(ax + b)^5$ given that the second term of its expansion is $240x^4$ and the third term…

Determine the positive real values of the parameters $a$ and $b$ in the binomial $(ax + b)^5$ given that the second term of its expansion is $240x^4$ and the third term of its expansion is $720x^3$.

Solve

Product of two expansions

To find a coefficient in a product like $(1+x)^4(1-2x)^5$ you don't need to multiply everything out: expand each binomial and add up only the products of terms whose exponents give the one you want.

To get x⁵ in a product, combine terms whose exponents add up to 5. (1+x)⁴ · (1−2x)⁵: pick x^i from the first and x^j from the second with i + j = 5, i from 0 to 4: (0,5) (1,4) (2,3) (3,2) (4,1)
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Calculate the coefficient of $x^5$ in the full expansion of the product of polynomials $(1 + x)^4 \times (1 - 2x)^5$.

Calculate the coefficient of $x^5$ in the full expansion of the product of polynomials $(1 + x)^4 \times (1 - 2x)^5$.

Solve

Properties and approximations

The sum of all the coefficients in a row is $2^n$ (obtained with $a=b=1$). And when one of the summands is very small, the first few terms are enough to approximate.

The sum of the coefficients of row n is equal to 2 to the power n. (1+1)ⁿ = sum of row n= 2ⁿ
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Prove that the sum of all binomial coefficients in a given row $n$ satisfies $\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$ using the expansion of the…

Prove that the sum of all binomial coefficients in a given row $n$ satisfies $\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n} = 2^n$ using the expansion of the binomial theorem for $(1+1)^n$. Apply this result to determine how many different groups (of any size, including the empty group) can be formed with the $8$ employees of an office.

Solve
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Calculate the sum of all the coefficients of the polynomial resulting from expanding $(3x - 2y)^5$. Hint: Think about what numerical values you can assign to the…

Calculate the sum of all the coefficients of the polynomial resulting from expanding $(3x - 2y)^5$. Hint: Think about what numerical values you can assign to the variables $x$ and $y$ so that the evaluated expression is exactly equal to the sum of its coefficients.

Solve
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In economics, the final compound capital accumulated after $5$ years at an annual compound interest rate of $r$ is calculated using the formula $C = C_0(1 + r)^5$. Use…

In economics, the final compound capital accumulated after $5$ years at an annual compound interest rate of $r$ is calculated using the formula $C = C_0(1 + r)^5$. Use the binomial theorem to expand $(1 + r)^5$. If the interest rate is $r = 0.03$, calculate an approximation of the final capital for an initial capital of $20\,000$ pesos using only the first three terms of the expansion.

Solve

Applications: counting and chance

When something is repeated $n$ times with two possible outcomes, the term $\binom{n}{k}p^{k}q^{n-k}$ in the expansion of $(p+q)^n$ groups together the cases with exactly $k$ "successes." The coefficient counts how many paths lead there.

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The volume of a metal cube of edge length $L$ at a certain temperature is $V = L^3$. When heated, its edge expands linearly to a length of $L(1 + \alpha \Delta T)$.…

The volume of a metal cube of edge length $L$ at a certain temperature is $V = L^3$. When heated, its edge expands linearly to a length of $L(1 + \alpha \Delta T)$. Expand the expression for the new volume using the binomial theorem and, knowing that the product $\alpha \Delta T$ is a very small value (of the order of $10^{-5}$), mathematically justify why the volume increase $\Delta V$ can be approximated by the expression $3 L^3 \alpha \Delta T$.

Solve
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A Galton board (or quincunx) is used to demonstrate the binomial distribution. A falling ball has the same probability of going right ($D$) as left ($I$) at each peg.…

A Galton board (or quincunx) is used to demonstrate the binomial distribution. A falling ball has the same probability of going right ($D$) as left ($I$) at each peg. The board in the figure has $4$ rows of pegs, so the ball makes $4$ moves and can land in $5$ bins. Each bin is labeled with the number of moves to the right that the ball has made. The possible paths are modeled by the expansion of $(I + D)^4$.

Galton board with 4 rows of pegs and 5 exit bins, with the fourth bin highlighted in red and an example path 0 D 1 D 2 D 3 D 4 D

Look at the figure: the dashed blue line shows a possible path to the bin highlighted in red, which corresponds to $1$ move to the left and $3$ to the right.

  1. a) Expand the binomial $(I + D)^4$.
  1. b) State algebraically how many different paths lead to the bin highlighted in red.
Solve
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In quality control at an electronic component factory, the probability that a batch of $n$ components has exactly $k$ defective items is given by the general term…

In quality control at an electronic component factory, the probability that a batch of $n$ components has exactly $k$ defective items is given by the general term $\binom{n}{k} p^k (1-p)^{n-k}$, where $p$ is the probability of an individual component being defective. If in a batch of $10$ components, the probability of finding exactly $4$ defective items is twice that of finding $3$ defective items, set up the corresponding equation and find the exact value of $p$.

Solve

Back to the beginning: $(2x-y)^4 = (2x)^4 - 4(2x)^3y + 6(2x)^2y^2 - 4(2x)y^3 + y^4 = 16x^4 - 32x^3y + 24x^2y^2 - 8xy^3 + y^4$, in a single step and without multiplying parentheses.

Summary

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