Choose a binomial expansion

Derive the binomial expansion

Let n be a constant and $f(x)=(1+x)^n$. Both source copies print the following complete derivative sequence and values at zero:

$$f(x)=(1+x)^n,\qquad f(0)=1.$$ $$f^{\prime}(x)=n(1+x)^{n-1},\qquad f^{\prime}(0)=n.$$ $$f^{\prime\prime}(x)=n(n-1)(1+x)^{n-2},\qquad f^{\prime\prime}(0)=n(n-1).$$ $$f^{\prime\prime\prime}(x)=n(n-1)(n-2)(1+x)^{n-3},$$ $$f^{\prime\prime\prime}(0)=n(n-1)(n-2).$$ $$f^{(4)}(x)=n(n-1)(n-2)(n-3)(1+x)^{n-4},$$ $$f^{(4)}(0)=n(n-1)(n-2)(n-3).$$

In general, for a positive integer r,

$$f^{(r)}(x)=n(n-1)\cdots(n-r+1)(1+x)^{n-r},$$ $$f^{(r)}(0)=n(n-1)\cdots(n-r+1).$$

Substitution into the Maclaurin formula gives the source’s MF27 expansion:

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ $$\hspace{1em}+\frac{n(n-1)\cdots(n-r+1)}{r!}x^r+\cdots.$$

The source states the representation range $-1<x<1$ (also written $|x|<1$ in its summary). Independently, this is a sufficient interval for general real n; endpoint behaviour depends on n and cannot be decided by copying the logarithm’s endpoint rule. The positive-integer case below terminates and has no such restriction.

Positive integer powers give a finite binomial theorem

For a positive integer n, the source gives

$$(a+b)^n=a^n+\binom n1 a^{n-1}b+\binom n2 a^{n-2}b^2+\cdots$$ $$\hspace{1em}+\binom nr a^{n-r}b^r+\cdots+\binom n{n-1}ab^{n-1}+b^n.$$

For integers r with $0\le r\le n$, the binomial coefficient is

$$\binom nr={}^{n}C_r=\frac{n!}{(n-r)!\,r!}.$$

The source identifies this formula as being in MF27. The resulting expansion is finite, with n+1 terms before numerical specialisations or cancellation. The polynomial identity holds for all real a and b; it is not confined by an infinite series’ convergence interval.

The printed factoring remark is

$$(a+b)^n=a^n\left(1+\frac ba\right)^n.$$

This lets an expansion in the form $(1+x)^n$ generate one for a general binomial by setting $x=b/a$. Independently, this rewriting requires $a\ne0$. If $a=0$ for positive integer n, use the original finite identity directly. For a nonterminating binomial expansion, transform the sufficient condition to $|b/a|<1$ as well; the factoring step alone does not remove it.

The source asks: What is the position of the term with binomial coefficient $\binom nr$? The Student leaves the answer blank; the Teacher gives the (r+1)th term. Counting starts with $a^n$ at r=0 as the first term, so $T_{r+1}=\binom nr a^{n-r}b^r$. Do not confuse the exponent index r with the term’s position.

Choose the expansion form and the required powers

1. Decide whether the exponent makes the expansion finite.
2. For the standard infinite form, factor the bracket so its constant term is 1.
3. Name the substituted argument and transform its validity condition.
4. List the powers needed in the final answer before expanding.

The source compares finite binomial expansion with the general binomial Maclaurin form in three rows:

Feature: Basic form

Finite binomial expansion: $(a+b)^n$

General binomial Maclaurin form: $(1+x)^n$; the bracket’s constant term must be 1.

Feature: Power n

Finite binomial expansion: Positive integer $n\in\mathbb Z^+$

General binomial Maclaurin form: n is a constant.

Feature: Number of terms

Finite binomial expansion: Finite, n+1 terms

General binomial Maclaurin form: The source says infinitely many terms.

The last entry describes a nonterminating binomial series; independently, a nonnegative integer exponent makes the Maclaurin expansion terminate too. These are overlapping cases, not mutually exclusive types of function. The requirement to begin with 1 concerns the bracket used for the standard binomial formula, not every Maclaurin series: sine, for example, has constant term 0.

Example 3 applies the two source callouts: factor $(a+bx)^n$ to $a^n(1+bx/a)^n$ when its bracket does not start with 1, then transform the standard argument condition to $|bx/a|<1$. The rewriting requires a nonzero a and suitable domain for fractional real powers.

It retains all the source’s cube-root coefficients and both inverse-expression methods. Its cube-root part repeats the expansion in Discussion question 6(i); that existing problem’s further approximation asks remain intact.

For $(1-1/x)^{-1}$, directly substituting $-1/x$ gives descending powers and a representation away from zero. Rewriting as $-x(1-x)^{-1}$ produces the requested ascending powers for $|x|<1$, while preserving the original $x\ne0$ condition.

Self-Practice 1 then contrasts a finite fourth power, an inverse-square expression and a negative fractional power. Its printed fourth-degree answer for part (a) is retained separately from the requested cubic truncation.

Use a nearby number without claiming an accuracy guarantee

Use the binomial expansion through the quadratic term to estimate $1/\sqrt{1.02}$. State the substituted small quantity and convergence condition. Then compare with a calculator value and distinguish displaying five decimal places from establishing five-decimal-place accuracy.

After your attempt: compare the reasoning

With $u=0.02$, the standard expansion gives

$(1+u)^{-1/2}=1-\frac u2+\frac{3u^2}{8}+\cdots,\qquad |u|<1.$

The quadratic estimate is $1-0.01+0.00015=0.99015$. The calculator value is approximately $0.990147543$, so in this case the estimate agrees when rounded to five decimal places. That numerical comparison checks this particular input; neither convergence nor printing five decimals alone proves that a truncation has that accuracy at other inputs.