The Binomial theorem follows from the expansion of (a+b)n. 1
(a+b)1 = a + b Notice that the coefficients 1 1
(a+b)² = a²+ 2ab + b² form a pattern that 1 2 1
a+b)³ = a ³+ 3a²b + 3ab² + b³ you have seen before 1 3 3 1
(a+b)⁴ = a⁴+ 4a³b + 6a²b²+4ab³ + b⁴ thanks to M.Pascal 1 4 6 4 1
Each number is the sum of the two numbers immediately above it. This is Pascal’s Triangle. The coefficients can be found on your calculator by using the nCr button, sometimes written nCr or nCr or as a vertical pair in brackets. On my calculator I press nCr, so 5C2 = 10, 17C6=12376.
A formula for nCr can be found, best represented using factorials. nCr = n! / r! (n-r)!
3! = 3.2.1=6, 5!=120, labelled n! or x! on your calculator. Check 60! (very large) and try to find 70!.
Textbooks often put the n high and the r low, as in nCr. I didn’t on this page because I want you to see the coefficients clearly.
Exercise:
1 Label the first four diagonals of Pascal’s triangle. The sequences have names.
2 Find the value of 20C15 - 20C5. Find 15C0 - 5C0. Explain both results.
3 As directed above, Find 70! and hence write 75! to 6 sig fig in standard form
This is still lower school maths. Look again at the first few lines; these are called expansions of the brackets. Note that the index of a goes down steadily and the index of b goes up correspondingly, so that the sum of the index is always the same as the order of the problem. For example, here’s a hard one: (x+y)17 = x17y⁰+ 17x17y1 + 136 x16y1 + 680 x15y² + 2380 x14y³ + 6188 x13y⁴ + 12376 x12y⁵ +..... and I’ve written in the y0 (y0 =1) and y1 (y1 =y) just to underline the point about index sum.
Exercise: textbook exercises are often written with x and y , so I have too.
4 Expand (x+y)10 completely.
5 Expand (x+y)20 to the fifth term.
6 Expand (x+y)30 to the fifth term
7 Expand (s + t)12 to the fifth term
Practical examples insert a number or two. Using dot for multiply, here’s a few examples:
(x+5)³ = x³ + 3.x².5 + 3.x.5² + 5³ = x³ + 15x² + 75x+ 125
(2x-5)³ = (2x)³ + 3.(2x)².(-5) + 3.2x.(-5)² + (-5)³ = 8x³- 60x² + 150x - 125
(3x+7)⁴ = (3x)⁴ + 4.(3x)³.7 + 6.(3x)².7² + 4.3x.7³ + 7⁴ = 81x⁴ + 756x³ + 2646x² + 4116x+ 2401
Now we’re getting out of lower school maths. At lower school level you should be near perfect at expanding brackets for quadratics and be able to do some cubics. Only accelerated and advanced sets would explore higher orders (that is (IMHO) what extension work should be; harder, but no new theory).
Prove you’re fine at lower school expansion:
8 (3x + 7)(4x - 5)
9 (3x+7)²
Show you can do extension material:
10 (x - 3)³
11 (2x - 3) ³
12 (7x - 1) ⁴
Okay, now we can get to Year 11 and 12 material. Expand these:
13 (x - 1/x)⁵ Use x=4 to find (3¾)⁵ as a fraction in the form p / 1024 (1024 = 4⁵)
14 (x + 1/x)⁷ Use x=3 and add up your terms to check that (3⅓)⁷ is 10⁷ / 3⁷
15 (p+q)⁵ Use your result when p=q=½ to indicate the probability that spinning five coins will give (i) exactly three heads, (ii) at least three heads
16 Use your answer to Q15 and p = 1-q = ⅙ to find the probability that 5 fair dice will produce (i) three sixes and (ii) at least three sixes
Summarising, the general case studied so far is
(x+y)ⁿ = nC0 xⁿ + nC1 xn-1y + nC2 xn-2y² + nC3 xn-3y³ + nC4 xn-4y⁴ + ....... + nCn-r xryn-r +.. + nCn-1 xyn-1 + yⁿ
with it understood that nCr = n! / r! (n-r)!
DJS 20130214
Link direct to Binomial Expansions 2 in Sixth Form Maths
Link direct to Binomial Distribution in Stats
1 Ones, counting numbers, triangle numbers, pyramid numbers.
2 it’s a symmetrical pattern, so of course they’re equal. nC0 = 1 for all n. nCn=1, too.
3 2.48091x10103
8 12x2 + (14-15)x - 35
11 8x3 - 36x2 + 54x - 27
12 2401x4 + 1372x3 + 294x2 + 28x+ 1
13 x5 - 5x3 + 10x - 10/x + 5/x3 - 1 /x5
14 x7 + 7x5 + 21x3 + 35x + 35/x + 21/x3 + 7/x5 + 1 /x7
15 P(3H) = 5C2 / 25 = 10/32 = 5/16 P(≥3H) = (5C0+5C1+5C2)/25 = 16/32 = 1/2
16 P(3S) = 5C2 52/65 = 250/7776 = P(≥3S) = (5C2*52 +5C1* 5 +5C0)/65 = 276/7776
