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Reference Sheet: Programme F.7 — Binomials

Rules, procedures and drill, in the order of the book's programme. Every rule links to the unit in the lesson that derives it — if a rule here looks arbitrary, the link is where it stops being arbitrary.

1. Factorials

From Unit 01.

n distinct items, arranged in order, have

n!=n(n-1)(n-2)\cdots2\cdot1

arrangements. By calculator, enter n and press the x! key. Defined so that the recursion n!=n(n-1)! holds at n=1: 0!=1.

Examples. 5!=120; 9!=362\,880; 11!=39\,916\,800.

2. Factorial manipulation

From Unit 01.

Peel a factorial back to a common tail and cancel it; never expand both factorials in full first.

\frac{(n+1)!}{(n-1)!}=n(n+1),\qquad\frac{(2n+1)!}{(2n-1)!}=(2n+1)(2n).

2n! is not (2n)!. 2n!=2\times(n!), doubling the factorial; (2n)! is the factorial of the doubled number.

Examples. \frac{12!}{9!}=12\times11\times10=1320. At n=3: 2n!=2\times6=12 but (2n)!=6!=720. At n=5: \frac{(2n)!}{(2n-2)!}=\frac{10!}{8!}=10\times9=90=2n(2n-1).

3. Combinations: the combinatorial coefficient

From Unit 02.

Placing r identical items in n distinct locations (0\le r\le n):

\boxed{{}^{n}C_{r}=\frac{n!}{(n-r)!\,r!}}

(also written \binom nr). Start from the ordered count \frac{n!}{(n-r)!} and divide by the r! orderings that look the same once the items are identical. The book calls this the combinatorial coefficient; arranging (also called permuting) is a different, larger count that keeps the order.

Examples. {}^{8}C_{3}=\frac{8!}{5!\,3!}=56; {}^{15}C_{9}=\frac{15!}{6!\,9!}=5005; {}^{4}C_{4}=1; {}^{3}C_{0}=1; {}^{5}C_{1}=5.

4. Three properties of combinatorial coefficients

From Unit 02, where each is proved both by counting and from the formula.

  1. {}^{n}C_{n}={}^{n}C_{0}=1 — using 0!=1.
  2. {}^{n}C_{n-r}={}^{n}C_{r} — choosing the r filled locations is the same act as choosing the n-r empty ones; in the formula, n-(n-r)=r swaps the two denominator factorials.
  3. Pascal's rule: {}^{n}C_{r}+{}^{n}C_{r+1}={}^{n+1}C_{r+1} — single out one location: left empty, or filled.

Examples. {}^{10}C_{4}={}^{10}C_{6}=210 (property 2, n=10,r=6). {}^{7}C_{3}+{}^{7}C_{4}=35+35=70={}^{8}C_{4} (property 3, n=7,r=3).

5. Pascal's triangle

From Unit 02.

A table of {}^{n}C_{r}, laid out left-justified as the book does, with row n and column r both numbered from 0. Every interior entry is the one directly above plus the one above-and-left (property 3); every edge is 1 (property 1). The lesson draws the same numbers centred, where the two parents sit diagonally above.

\begin{array}{c|cccccc} n\backslash r & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 1 & & & & & \\ 1 & 1 & 1 & & & & \\ 2 & 1 & 2 & 1 & & & \\ 3 & 1 & 3 & 3 & 1 & & \\ 4 & 1 & 4 & 6 & 4 & 1 & \\ 5 & 1 & 5 & 10 & 10 & 5 & 1 \end{array}

Example. Row 5, column 2: 10=6+4, the entry directly above ({}^{4}C_{2}=6) plus the one above-and-left ({}^{4}C_{1}=4), which is {}^{5}C_{2}={}^{4}C_{2}+{}^{4}C_{1}.

6. Binomial expansions

From Unit 03.

(a+b)^n=\sum_{r=0}^{n}{}^{n}C_{r}\,a^{n-r}b^{r}=a^n+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+\frac{n(n-1)(n-2)}{3!}a^{n-3}b^3+\cdots+b^n.

Row n of Pascal's triangle gives the coefficients directly for small n; the \frac{n(n-1)\cdots}{r!} form is faster for large n. The two exponents in every term add to n.

Examples. (a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4. Row 7 gives (1+x)^7=1+7x+21x^2+35x^3+35x^4+21x^5+7x^6+x^7.

7. Expansions with a negative or fractional second term

From Unit 03.

Rewrite (a-c)^n as \big(a+[-c]\big)^n and substitute b=-c throughout, so every b^r carries the sign (-1)^r automatically. The same substitution handles descending powers of x, where b is a fraction such as -\frac3x.

Example. (3-2x)^4=\big(3+[-2x]\big)^4, with a=3, b=-2x and row 4 (1,4,6,4,1):

(3-2x)^4=3^4+4(3^3)(-2x)+6(3^2)(-2x)^2+4(3)(-2x)^3+(-2x)^4
=81+4(27)(-2x)+6(9)(4x^2)+4(3)(-8x^3)+16x^4=81-216x+216x^2-96x^3+16x^4.

8. The general term of the binomial expansion

From Unit 03.

The (r+1)th term, 0\le r\le n — r is one less than the term number:

T_{r+1}={}^{n}C_{r}\,a^{n-r}b^{r}.

"Ascending powers of x" fixes which symbol is b (the one carrying x to a rising power); for descending powers the same general term applies with b a fraction in x.

Examples. The 10th term of (1+x)^{15}, ascending, has r=9: T_{10}={}^{15}C_{9}x^9=5005x^9. The 8th term of \left(2-\frac x3\right)^{12} has r=7: T_8={}^{12}C_7\,2^5\left(-\frac x3\right)^7=-\frac{2816}{243}x^7. The 4th term of \left(2-\frac3x\right)^8, descending, has r=3: T_4={}^{8}C_3\,2^5(-3)^3x^{-3}=-\frac{48\,384}{x^3}; the coefficient of x^{-4} has r=4: {}^{8}C_4\,2^4(-3)^4=90\,720.

9. Sigma notation for the binomial expansion; row sums

From Unit 02, Unit 03 and Unit 04.

(a+b)^n=\sum_{r=0}^{n}{}^{n}C_{r}\,a^{n-r}b^{r}.

Setting a=b=1 turns every term into {}^{n}C_{r}, so the sum of row n of Pascal's triangle is

\sum_{r=0}^{n}{}^{n}C_{r}=2^n.

Example. Row 4: 1+4+6+4+1=16=2^4.

10. Forming general terms

From Unit 04.

Pattern General term
Even integers 2r
Odd integers, starting 1 2r-1
Odd integers, starting 3 2r+1
Alternating, + on even r (-1)^r\,(\ldots)
Alternating, + on odd r (-1)^{r+1}\,(\ldots)
Counting starts at r=0 the nth term sits at r=n-1

Examples. 16=2(8), so 16 is the 8th even term. 21=2(11)-1, the 11th odd term. The series -1+2-3+4-\cdots has general term (-1)^r r.

11. The sum of the first n natural numbers

From Unit 04.

\sum_{r=1}^{n}r=\frac{n(n+1)}2

by pairing the sum with its reverse: every one of the n columns adds to n+1.

Examples. \sum_{r=1}^{50}r=1275; \sum_{r=1}^{30}r=465.

12. Rules for manipulating sums

From Unit 04.

\textbf{Rule 1: }\sum_{r=1}^{n}k\,f(r)=k\sum_{r=1}^{n}f(r)\ \ (\text{so }\textstyle\sum_{r=1}^n k=kn),\qquad\textbf{Rule 2: }\sum_{r=1}^{n}\big(f(r)+g(r)\big)=\sum_{r=1}^{n}f(r)+\sum_{r=1}^{n}g(r).

Combined with §11, any linear general term sums in closed form.

Examples. \sum_{r=1}^{n}(6r+5)=6\cdot\frac{n(n+1)}2+5n=n(3n+8); at n=2: 28=11+17. \sum_{r=1}^{n}(8r-7)=8\cdot\frac{n(n+1)}2-7n=n(4n-3); at n=4: 52=1+9+17+25. \sum_{r=1}^{n}(4r+2)=2n(n+2); at n=3: 30=6+10+14.

13. Beyond the book: \sum r^2 and \sum r^3

From Unit 04.

Not part of programme F.7 (the book uses \sum r^2 only as a notation example), but useful enrichment. The staircase pairing of §11 does not extend to squares; these come from telescoping. Summing (r+1)^3-r^3=3r^2+3r+1 over r=1,\ldots,n collapses the left side to (n+1)^3-1, which is then solved for \sum r^2; summing (r+1)^4-r^4=4r^3+6r^2+4r+1 does the same for \sum r^3. Both are worked in full in the lesson.

\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}6,\qquad\sum_{r=1}^{n}r^3=\left[\frac{n(n+1)}2\right]^2.

Example, n=5. \sum r=15; \sum r^2=55=\frac{5\cdot6\cdot11}6; \sum r^3=225=15^2. In general \sum r^3=\left(\sum r\right)^2, because the derivation gives both as \left[\frac{n(n+1)}2\right]^2.

14. The exponential number e

From Unit 05.

e=\lim_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{r=0}^{\infty}\frac1{r!}=2.7182818\ldots

The two forms are the same number. Expanding \left(1+\frac1n\right)^n binomially, term r is \frac1{r!} times bracket factors \left(1-\frac kn\right) that climb toward 1 as n grows, so the expansion rises with n and never exceeds \sum\frac1{r!}. That sum is finite: r!\ge2^{r-1}, so \frac1{r!}\le\frac1{2^{r-1}} and every running total is below 1+2=3. Keeping a fixed number of terms and letting n\to\infty shows the limit is also at least every running total, so the two agree.

Examples. \left(1+\frac1n\right)^n at n=1,2,12,365: 2, 2.25, 2.613, 2.7146. Six terms of the series: 1+1+\frac12+\frac16+\frac1{24}+\frac1{120}=2.7166667.

15. The series for e^x and the stopping rule

From Unit 05.

e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots

Stopping rule. Add terms until the next term is too small to change the required decimal places or significant figures. When the terms alternate in sign and shrink (as they do for -1<x<0), consecutive running totals straddle the true value, so once two consecutive totals round to the same answer, that answer is certain.

Examples. e^{0.1} to 3 sig fig: 1+0.1+0.005+0.000167=1.105167, and the next term is 0.0000042, so e^{0.1}\approx1.11. e^{0.2} to 4 dp: 1+0.2+0.02+0.001333+0.0000667+0.0000027=1.2214024, so e^{0.2}\approx1.2214. e^{-0.25} to 3 dp: 1-0.25+0.03125-0.0026042+0.0001628-0.0000081=0.7788005; the consecutive totals 0.7786458 and 0.7788086 trap the true value and both round to 0.779, so e^{-0.25}\approx0.779.

Traps

  • Forgetting 0!=1. \frac{5!}{5!\,0!} is 1, not undefined or 0 — needed for {}^{n}C_0={}^{n}C_n=1.
  • 2n! read as (2n)!. At n=3 these are 12 and 720 — not remotely close.
  • Permutations used where combinations are wanted. 8\times7\times6=336 counts ordered placements; identical items need the \div r! step, giving 56.
  • Off-by-one in the general term. T_{r+1} uses r, so the "5th term" is r=4, not r=5.
  • Losing the sign on a negative second term. In (3-2x)^4 every odd-r term is negative because b=-2x; only b carries the sign, never a.
  • Mixing up ascending and descending powers. For \left(2-\frac3x\right)^8, "descending powers of x" still means b=-\frac3x in the same general term — the power of x falls because b itself contains x^{-1}, not because the formula changes.
  • Wrong starting index. A series stated with r from 0 has an nth term at r=n-1, one less than a series starting at r=1.
  • Factoring out only part of a constant. \sum(6r+5) needs both the 6 pulled out by Rule 1 and the constant 5 handled by \sum_{r=1}^n 5=5n, not left as "+5" once.
  • Stopping the e^x series too soon. Check the size of the next term against the required accuracy before stopping, not just how many terms "feels enough".
  • Treating \sum r^2 and \sum r^3 as examinable. They are enrichment (§13); the programme's only closed-form power sum is \sum r.

Self-check

  1. Seven distinct inspection tasks are run one after another, each once. In how many orders can they be run? Also evaluate \frac{16!}{13!}.
  2. Simplify \frac{(n+2)!}{n!} and check it at n=4. Then evaluate 2n! and (2n)! at n=4.
  3. Six identical spacers are fitted into a jig with 9 numbered slots, at most one per slot. In how many ways can this be done? Explain why the answer equals {}^{9}C_{3}.
  4. Verify Pascal's rule for n=6, r=2 from the formula, then build row 6 of Pascal's triangle from row 5.
  5. Expand (2a-b)^4.
  6. Find the 5th term of (1-2x)^8 in ascending powers of x.
  7. Find the coefficient of x^{-3} in \left(3-\frac2x\right)^7, expanded in descending powers of x.
  8. Show that \sum_{r=0}^{4}(-1)^r\,{}^{4}C_r=0.
  9. Write the first n terms of 5-10+15-20+\cdots and of 1-2+4-8+\cdots in sigma notation, and add the first 6 terms of the first series.
  10. Find the 6th term and the sum of the first 15 terms of 4+8+12+16+\cdots.
  11. Evaluate \sum_{r=1}^{6}(3r-2) directly, then find the closed form in n and check it at n=6.
  12. Evaluate e^{0.3} to 4 dp using the series and its stopping rule.

Fully worked solutions

1. Each task is used once, so the choices shrink by one at each step: 7!=7\times6\times5\times4\times3\times2\times1=5040 orders. \frac{16!}{13!}=\frac{16\times15\times14\times13!}{13!}=16\times15\times14=3360.

2. (n+2)!=(n+2)(n+1)\,n!, so \frac{(n+2)!}{n!}=(n+2)(n+1). At n=4: \frac{6!}{4!}=\frac{720}{24}=30 and (4+2)(4+1)=6\times5=30. ✓ Then 2n!=2\times4!=2\times24=48, while (2n)!=8!=40\,320.

3. Identical items in distinct locations: {}^{9}C_{6}=\frac{9!}{3!\,6!}=\frac{9\times8\times7}{3\times2\times1}=\frac{504}{6}=84 ways. Choosing the 6 filled slots is the same act as choosing the 3 empty ones, so {}^{9}C_{6}={}^{9}C_{3}=84 (property 2).

4. {}^{6}C_{2}=\frac{6!}{4!\,2!}=15 and {}^{6}C_{3}=\frac{6!}{3!\,3!}=20, so {}^{6}C_{2}+{}^{6}C_{3}=35; and {}^{7}C_{3}=\frac{7!}{4!\,3!}=35. ✓ Row 6 from row 5 (1,5,10,10,5,1), each interior entry the sum of the two above it: 1,\ 1+5=6,\ 5+10=15,\ 10+10=20,\ 10+5=15,\ 5+1=6,\ 1.

5. (2a-b)^4=\big(2a+[-b]\big)^4 with row 4 (1,4,6,4,1): (2a)^4+4(2a)^3(-b)+6(2a)^2(-b)^2+4(2a)(-b)^3+(-b)^4=16a^4-32a^3b+24a^2b^2-8ab^3+b^4.

6. The 5th term has r=4, with a=1, b=-2x, n=8: T_5={}^{8}C_4\,1^4(-2x)^4=70\times16x^4=1120x^4.

7. a=3, b=-\frac2x, n=7. Since b^r carries x^{-r}, the term in x^{-3} has r=3: T_4={}^{7}C_3\,3^4\left(-\frac2x\right)^3=35\times81\times\left(-\frac8{x^3}\right)=-\frac{22\,680}{x^3}, so the coefficient is -22\,680.

8. \sum_{r=0}^{4}(-1)^r\,{}^{4}C_r=1-4+6-4+1=0. ✓ (Putting a=1,b=-1 into the binomial theorem gives (1-1)^n=0 for every n>0, which is exactly this sum.)

9. First series: sizes 5r, sign positive on odd r, so \sum_{r=1}^{n}(-1)^{r+1}5r. Second series: sizes 2^r starting from 2^0=1 at r=0, positive on even r, so \sum_{r=0}^{n-1}(-1)^r2^r (counting from r=0, the nth term sits at r=n-1). First 6 terms of the first series, in pairs: (5-10)+(15-20)+(25-30)=3\times(-5)=-15.

10. General term 4r; 6th term 4\times6=24. Sum of the first 15 terms: 4\sum_{r=1}^{15}r=4\times\frac{15\times16}2=4\times120=480.

11. Directly: 1+4+7+10+13+16=51. Closed form: \sum_{r=1}^n(3r-2)=3\cdot\frac{n(n+1)}2-2n=\frac{3n^2+3n-4n}2=\frac{n(3n-1)}2; at n=6: \frac{6\times17}2=51. ✓

12. With x=0.3 the terms are 1, 0.3, \frac{0.09}{2}=0.045, \frac{0.027}{6}=0.0045, \frac{0.0081}{24}=0.0003375, \frac{0.00243}{120}=0.00002025; running total 1.34985775. The next term is \frac{0.000729}{720}=0.0000010, and it and all later terms together cannot lift the total to 1.34995, so e^{0.3}\approx1.3499.