Title HSC Maths Extension 1 Notes 39.7 KB 5
##### Document Text Contents
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The numbers in Pascal’s triangle are given by the nCr formula. In fact, we can write

(1 + x) = 1C0 +
1C1x

(1 + x)2 = 2C0 +
2C1x+

2C2x
2

(1 + x)3 = 3C0 +
3C1x+

3C2x
2 + 3C3x

3

(1 + x)4 = 4C0 +
4C1x+

4C2x
2 + 4C3x

3 + 4C4x
4

and so on. The general form is

(1 + x)n = nC0 +
nC1x+

nC2x
2 + ...+ nCnx

n

and more generally again, we have

(a+ b)n = nC0a
n + nC1a

n−1b+ nC2a
n−2b2 + ...+ nCnb

n

Generally, when expanding out we use Pascal’s triangle rather than the nCr formula, but
the formula is very useful in doing other more theoretical problems. The general term in the
expansion of (a+ b)n is

nCra
n−rbr =

(
n
r

)
an−rbr.

Ex: Find the co-efficient of x8 and the constant term in the expansion of (2x3 − 1
x
)12.

The general term is

(
12
r

)
(2x3)12−r × (− 1

x
)r =

(
12
r

)
212−r × x36−4r × (−1)r. Now

36− 4r = 8 when r = 7 and so the coefficient of x8 is
(

12
7

)
25 × (−1)7 = −25344.

The constant term will appear when 36− 4r = 0 and so r = 9. Hence the constant term is(
12
9

)
23 × (−1)9 = −1760.

Integration by Substitution.

Many integrals can be performed by making a change of variable.

Ex. Find I =

xex

2

dx, using u = x2.

Put u = x2, then du
dx

= 2x. Hence we can symbolically replace x dx by du
2
. The integral

then becomes

I =
1

2

eu du =

1

2
eu + C =

1

2
ex

2

+ C.

Harder Inequalities.