Series Expansions using the Binomial Theorem

The Binomial theorem follows from the expansion of (a+b)                                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 = a³ + 3a²b + 3ab² + b³                 you have seen before                   1  3  3  1

(a+b)= a+ 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 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.


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 os 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 =1) 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.(-5² + (-5)³  =  8x³ - 60x² + 150x - 125

(3x+7)⁴ =  (3x)+ 4.(3x)³.7 + 6.(3x)².72 + 4.3x.73 + 74   =  81x+ 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)

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 = 4)

14    (x + 1/x)⁷             Use x=3 and add up your terms to check that  (3 ⅓)⁷  is    10 / 3

15   (p+q)⁵               Use your result when p=q=1/2 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


I’ll put this page in Lower School.


Summarising, the general case studied so far is


(x+y) = nC0 x + nC1 xn-1y + nC2 xn-2y² + nC3 xn-3y³ + nC4 xn-4y+ ....... + nCn-r xryn-r +.. + nCn-1 xyn-1 + y


   with it understood that      nCr = n! / r! (n-r)!


DJS 20130214

In statistics, probability calculations are based upon expansions of (p+q), where p+q = 1. Many students get stuck here, seeing that 1 is obviously one and therefore, they think, of no interest. Au contraire, mon brave; adding up to one simply confirms that the probabilities are all there and it is the way the ‘one’ is divided up that tells you the distribution of the various probabilities. That’s the interesting bit, the distribution.

I put an example in the Lower School Binomial. The usage is in Hypothesis testing. there I have assumed you recognise “X is a variate that behaves as a binomial distribution with n items of probability p” as being written X~B(n,p) and that the mean is np (that’s the expectation) and the variance is npq. I have not proved that on these pages. Yet.


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  2401x+ 1372x³ + 294x² + 28x+ 1

13  x- 5x³ + 10x - 10/x + 5/x³ - 1 /x

14  x+ 7x+ 21x³ + 35x + 35/x + 21/x³ + 7/x⁵  + 1 /x

15  P(3H) = 5C2 / 2= 10/32 = 5/16     P(≥3H) = (5C0+5C1+5C2)/2= 16/32 = 1/2

16  P(3S) = 5C2  5²/6= 250/7776 =    P(≥3S) = (5C2*5² +5C1* 5 +5C0)/6= 276/7776

 

© David Scoins 2017