(i) Find the first three terms, in ascending powers of x, in the expansion of
(a) \((1-x)^6\),
(b) \((1+2x)^6\).
(ii) Hence find the coefficient of \(x^2\) in the expansion of \([(1-x)(1+2x)]^6\).
Solution
(i) (a) The expansion of \((1-x)^6\) using the binomial theorem is:
\((1-x)^6 = 1 - 6x + 15x^2 + \ldots\)
(b) The expansion of \((1+2x)^6\) using the binomial theorem is:
\((1+2x)^6 = 1 + 12x + 60x^2 + \ldots\)
(ii) To find the coefficient of \(x^2\) in \([(1-x)(1+2x)]^6\), we consider the product of the expansions:
\((1-x)^6 = 1 - 6x + 15x^2\)
\((1+2x)^6 = 1 + 12x + 60x^2\)
The coefficient of \(x^2\) is obtained by considering the terms:
\(1 \cdot 60x^2 = 60x^2\)
\(-6x \cdot 12x = -72x^2\)
\(15x^2 \cdot 1 = 15x^2\)
Adding these gives the coefficient of \(x^2\):
\(60 - 72 + 15 = 3\)
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