Expert: Paul Klarreich - 2/5/2008

Question
a) Write down the general term in the power series expansion of (x^2 - 2/x)^18
Hence find the term which is independent of x.

b) Find the first four terms in the expansion of (1+x)^16 in ascending powers of x. Hence expand (1+z+z^2)^16 in ascending powers of z up to and including the term in z^3

Answer
Questioner:   Chloe
Category:  Advanced Math
Private:  No
 
Subject:  Series
Question:  a) Write down the general term in the power series expansion of (x^2 - 2/x)^18
Hence find the term which is independent of x.

b) Find the first four terms in the expansion of (1+x)^16 in ascending powers of x. Hence expand (1+z+z^2)^16 in ascending powers of z up to and including the term in z^3
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Hi, Chloe,

This is not 'series' -- it's the Binomial expansion.

a) Write down the general term in the power series expansion of (x^2 - 2/x)^18
Hence find the term which is independent of x.

The general term in a Binomial expansion of (a + b)^n is  C(n,r)a^(n-r)b^r

Now put a = x^2, b = -2/x,  n = 18.  So your term is:

C(n,r)(x^2)^(18-r) (-2/x)^r

Now if you want one that doesn't have x, you need:  2(18 - r) = r.  Solve that:

36 - 2r = r
r = 12.

that term is:  C(18,12)(x^2)^6(-2/x)^12 = C(18,12)(-2)^12, which you can work out.
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b) Find the first four terms in the expansion of (1+x)^16 in ascending powers of x. Hence expand (1+z+z^2)^16 in ascending powers of z up to and including the term in z^3

(1 + x)^16 =  1 + 16x + 16*15/2 x^2 + (16*15*14/3*2*1) x^3 + ...

(1 + x)^16 =  1 + 16x + 120 x^2 + 560 x^3 + ...


To expand (1 + z + z^2)^16, you can do it several ways.  One could be:

(1 + (z + z^2))^16 and set  x = (z + z^2)

1 + 16(z + z^2) + 120(z + z^2)^2 + 560(z + z^2)^3

Now expand and just throw away (...) any terms that go above z^3:

1 + 16z + 16z^2 + 120(z^2 + 2z^3 + ...) + 560(z^3 + ........)

1 + 16z + 16z^2 + 120z^2 + 240z^3 + 560z^3

Now just combine those.