(i) Find the coefficients of \(x^4\) and \(x^5\) in the expansion of \((1 - 2x)^5\).
(ii) It is given that, when \((1 + px)(1 - 2x)^5\) is expanded, there is no term in \(x^5\). Find the value of the constant \(p\).
Solution
(i) The binomial expansion of \((1 - 2x)^5\) is given by:
\((1 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-2x)^k\)
For \(x^4\), \(k = 4\):
Coefficient = \(\binom{5}{4} (1)^{1} (-2)^4 = 5 \times 16 = 80\)
For \(x^5\), \(k = 5\):
Coefficient = \(\binom{5}{5} (1)^{0} (-2)^5 = 1 \times (-32) = -32\)
(ii) In the expansion of \((1 + px)(1 - 2x)^5\), the term in \(x^5\) is:
\((-32) + 5p(1)(-2)^4 = 0\)
\(-32 + 80p = 0\)
\(80p = 32\)
\(p = \frac{32}{80} = \frac{2}{5}\)
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