We pass now to the more general case of the expansion of (x + y)n. Because x and y are both variables, the symmetries of the expansion will be more obvious, and this section offers proofs of the addition property and the other basic patterns in the Pascal triangle.
The Pattern of the Indices in the Expansion of (x + y)n: Here are the expansions of (x + y)n for low values of n. Again, the calculations have been carried out with like terms written in the same column so that the addition property is clear.
(x + y)0 = 1 (x + y)1 =x + y
(x + y)2 =x(x + y) + y(x + y)
=x2+ xy + xy + y2
=x2+ 2xy + y2
(x + y)3 =x(x + y)2+y(x + y)2
=x3+ 2x2y + xy2 + x2y + 2xy2+y3
=x3+ 3x2y + 3xy2+y3 (x + y)4 =x(x + y)3+y(x + y)3
=x4+ 3x3y + 3x2y2+ xy3 + x3y + 3x2y2+ 3xy3+y4
=x4+ 4x3y + 6x2y2+ 4xy3+y4 The pattern for the indices of x and y is straightforward. The expansion of (x + y)3, for example, has four terms, and in each term the indices ofx and y are whole numbers adding to 3. Similarly the expansion of (x + y)4 has five terms, and in each term the indices of x and y are whole numbers adding to 4. The useful phrase for this is that (x + y)n is homogeneous of degree n in x and y together.
2
THE TERMS OF(x + y)n: The expansion of (x + y)n hasn + 1 terms, and in each term the indices of x and y are whole numbers adding to n.
That is, the expression (x + y)n is homogeneous of degree n in x and y together, and so also is its expansion.
r
180 CHAPTER5: The Binomial Theorem CAMBRIDGEMATHEMATICS3 UNITYEAR12r
The reason for this is clear, and formal proof by induction should not be necessary.
In each successive expansion, the terms of the previous expansion are multiplied first by x and then by y, so the sum of the indices goes up by 1. We now know that in general
(x + y)n =∗xn + ∗xn−1y + ∗xn−2y2 + · · · + ∗x2yn−2 + ∗xyn−1 + ∗yn, where∗ denotes the different coefficients.
A Symbol for the Coefficients: The coefficients here are, of course, the same as in the previous section, as can be seen by replacingx and y by 1 and x in the expansion of (x+y)n, and we will first deal with the special case of the expansion of (1+x)n. To investigate these coefficients further, we take the approach of giving names to the things we want to study.
3
THE DEFINITION OFnCr: Define the number nCr to be the coefficient of xr in the expansion of (1 +x)n.
The symbol is usually read as ‘n choose r’, and the notations nCr and n r
are both used for these coefficients.
This definition will need some thought. Defining a number as a coefficient in an expansion is standard practice in mathematics, but it seems very strange the first time it is encountered. We can now write out the expansion of (1 +x)n.
4
THE EXPANSION OF(1 +x)n: Using the notationnCr for the coefficients, (1 +x)n =nC0 + nC1x + nC2x2 + · · · + nCnxn. There aren + 1 terms, and the general term of the expansion is
term in xr =nCrxr.
Alternatively, using sigma notation, the expansion can be written as (1 +x)n =
n r=0
nCrxr.
WORKEDEXERCISE:
(a) Write out the expansion of (1 +x)2 and (1 +x)3 usingnCr notation.
(b) Hence give the values of 2C0,2C1,2C2 and of3C0,3C1,3C2,3C3. SOLUTION:
(a) (1 +x)2=2C0 + 2C1x + 2C2x2
and (1 +x)3=3C0 + 3C1x + 3C2x2 + 3C3x3 (b) But (1 +x)2= 1 + 2x + x2,
so 2C0= 1, 2C1 = 2 and 2C2= 1.
Also (1 +x)3= 1 + 3x + 3x2+x3,
so 3C0= 1, 3C1 = 3, 3C2 = 3 and 3C3 = 1.
r
CHAPTER5: The Binomial Theorem 5B Further Work with the Pascal Triangle 181r
The Expansion of (x + y)n: The coefficients in the expansion of (x + y)n are the same numbers nCr as in the expansion of (1 +x)n. Thus we can now write out the expansion of (x + y)n as well.
5
THE EXPANSION OF(x + y)n: Using the nCr notation,
(x + y)n =nC0xn + nC1xn−1y + nC2xn−2y2 + · · · + nCnyn. There are n + 1 terms, and the general term of the expansion is
term inxn−ryr =nCrxn−ryr.
Alternatively, using sigma notation, the expansion can be written as (x + y)n =
n r=0
nCrxn−ryr.
The Pascal Triangle: The Pascal triangle of the previous section now becomes the table of values of the functionnCr, with the rows indexed by n and the columns by r:
nCr 0 1 2 3 4 5 6 7 8
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
6 1 6 15 20 15 6 1
7 1 7 21 35 35 21 7 1
8 1 8 28 56 70 56 28 8 1
The boxed numbers provide another example of the addition property of the Pascal triangle, and will be discussed further below.
Using the General Expansion: The general expansion of (x+y)n is applied in the same way as the expansion of (1 +x)n.
WORKEDEXERCISE: Use the Pascal triangle to write out the expansions of:
(a) (2− 3x)4 (b) (5x +15a)5
SOLUTION:
(a) (2− 3x)4 = 24 + 4× 23× (−3x) + 6 × 22× (−3x)2 + 4× 2 × (−3x)3 + (−3x)4
= 16− 96x + 72x2− 216x3+ 81x4
(b) (5x +15a)5 = (5x)5 + 5× (5x)4×15a + 10 × (5x)3× (15a)2 + 10× (5x)2× (15a)3 + 5× (5x) × (15a)4+ (15a)5
= 3125x5+ 625ax4+ 50a2x3+ 2a3x2+251a4x +31251 a5
WORKEDEXERCISE: Use the Pascal triangle to write out the expansion of (2x + x−2)6, leaving the terms unsimplified. Hence find:
(a) the term independent ofx, (b) the term in x−3.
r
182 CHAPTER5: The Binomial Theorem CAMBRIDGEMATHEMATICS3 UNITYEAR12r
SOLUTION: (2x + x−2)6 = (2x)6 + 6× (2x)5× (x−2) + 15× (2x)4× (x−2)2
+ 20× (2x)3× (x−2)3 + 15× (2x)2× (x−2)4 + 6× (2x) × (x−2)5 + (x−2)6 (a) Constant term
= 15× (2x)4× (x−2)2
= 15× 24× x4× x−4
= 240.
(b) Term inx−3
= 20× (2x)3× (x−2)3
= 20× 23× x3× x−6
= 160.
WORKEDEXERCISE: Expand (2− 3x)7 as far as the term inx2, and hence find the term inx2 in the expansion of (5 +x)(2 − 3x)7.
SOLUTION: (2− 3x)7 = 27 − 7 × 26× (3x) + 21 × 25× (3x)2 − · · ·
= 128 − 1344 x + 6048 x2 − · · · . Hence the term inx2 in the expansion of (5 +x)(2 − 3x)7
= 5× 6048 x2 − x × 1344 x
= 28 896x2.
Proofs of the First Three Basic Properties: The first three basic properties of the Pascal triangle can now be expressed innCr notation and proven straightforwardly.
6
BASIC PROPERTIES OF THE PASCAL TRIANGLE: 1. Each row starts and ends with 1, that is,
nC0 =nCn = 1, for all cardinals n.
2. Each row is reversible, that is,
nCr =nCn−r, for all cardinalsn and r with r ≤ n.
3. The sum of each row is 2n, that is,
nC0+nC1+nC2+· · · +nCn = 2n, for all cardinalsn.
Proof: Each proof begins with the general expansion
(x + y)n =nC0xn + nC1xn−1y + nC2xn−2y2 + · · · + nCnyn. Parts 1 and 3 then proceed by substitution, and part 2 by equating coefficients.
These methods are both commonly required for solving problems, and they should be studied carefully.
1. Substitutingx = 1 and y = 0, (1 + 0)n =nC0+ 0 +· · · + 0, and so as required, 1 =nC0.
Substituting x = 0 and y = 1, (0 + 1)n = 0 + 0 +· · · + 0 +nCn and so as required, 1 =nCn.
2. We know that (x + y)n = (y + x)n.
Now (x + y)n =nC0xn + nC1xn−1y + · · · + nCn−1xyn−1 + nCnyn, and (y + x)n =nC0yn + nC1yn−1x + · · · + nCn−1yxn−1 + nCnxn. Equating coefficients of like terms in the two expansions,
nC0 =nCn, nC1 =nCn−1, nC2 =nCn−2, . . . , nCn =nC0, and in general nCn−r =nCr, forr = 0, 1, 2, . . . , n.
3. Substitutingx = 1 and y = 1, (1 + 1)n =nC0+nC1+nC2+· · · +nCn, and so as required, 2n =nC0+nC1+nC2+· · · +nCn.
r
CHAPTER5: The Binomial Theorem 5B Further Work with the Pascal Triangle 183r
Proof of the Addition Property of the Pascal Triangle: The addition property also needs to be restated innCr notation. In words, it says that every number in the triangle is the sum of the number directly above it, and the number above and to the left of it (apart from the first and the last numbers of each row). The boxed numbers in the Pascal triangle above provide an example of this — they show that
5C2=4C2+4C1 (that is, 10 = 6 + 4). The general statement, in symbolic form, is therefore:
7
THE ADDITION PROPERTY: Ifn and r are positive integers with r ≤ n, then
n+1Cr =nCr+nCr−1, for 1 ≤ r ≤ n.
Proof: The expansions at the start of Sections 5A and 5B were written so that the columns aligned to make the addition property obvious. A formal proof will require examination of the coefficients in the expansion of (1 +x)n+1. We begin by noticing that
Equating coefficients of these two terms proves the result.
Exercise 5B
Note: Questions 3 and 4 should be omitted by those wanting to delay the introduction ofnCr notation until Section 5D.
1. Use Pascal’s triangle to expand each of the following:
(a) (x + y)4
2. Use Pascal’s triangle to expand each of the following:
(a) (1 +x2)4
4. Use the values ofnCr from the Pascal triangle in the notes above to find:
(a) 6C0+6C2+6C4+6C6
(b) 6C1+6C3+6C5
(c) 2C2+3C2+4C2+5C2
(d) (5C0)2+ (5C1)2+ (5C2)2+ (5C3)2+ (5C4)2+ (5C5)2 5. Simplify the following without expanding the brackets:
(a) y5+ 5y4(x − y) + 10y3(x − y)2+ 10y2(x − y)3+ 5y(x − y)4+ (x − y)5 (b) a4− 4a3(a − b) + 6a2(a − b)2− 4a(a − b)3+ (a − b)4
r
184 CHAPTER5: The Binomial Theorem CAMBRIDGEMATHEMATICS3 UNITYEAR12r
(b) Similarly, differentiate x5 from first principles.10. (a) Show that (3 +√ a common denominator. Then simplify the expression using Pascal’s triangle.
(b) Similarly, simplify 1
13. Expand (x + 2y)5 and hence evaluate: (a) (1·02)5 correct to to five decimal places, (b) (0·98)5 correct to to five decimal places, (c) (2·2)5 correct to four significant figures.
14. (a) Expand: (i) 15. Find the coefficients of x and x−3 in the expansion of
r
CHAPTER5: The Binomial Theorem 5C Factorial Notation 185r
16. The coefficients of the terms in a3 and a−3 in the expansion of
ma + n a2
6
are equal, wherem and n are nonzero real numbers. Prove that m2:n2 = 10 : 3.
17. (a) Expand
x +1
x
6
. (b) IfU = x+1
x, expressx6+ 1
x6 in the formU6+AU4+BU2+C.
State the values ofA, B and C.
E X T E N S I O N
18. Find the term independent ofx in the expansion of (x + 1 + x−1)4. 19. [The Sierpinski triangle fractal]
(a) Draw an equilateral triangle of side length 1 unit on a piece of white paper. Join the midpoints of the sides of this triangle to form a smaller triangle. Colour it black.
Repeat this process on all white triangles that remain. What do you notice?
(b) Draw up Pascal’s triangle in the shape of an equilateral triangle, then colour all the even numbers black and leave the odd numbers white. What do you notice? This pattern will be more evident if you take at least the first 16 rows — perhaps use a computer program to generate 100 rows of Pascal’s triangle.