(i) Find the coefficients of \(x^2\) and \(x^3\) in the expansion of \((2-x)^6\).
(ii) Find the coefficient of \(x^3\) in the expansion of \((3x+1)(2-x)^6\).
Solution
(i) To find the coefficients of \(x^2\) and \(x^3\) in \((2-x)^6\), use the binomial theorem:
\((2-x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-x)^k\).
The coefficient of \(x^2\) is given by \(\binom{6}{2} (2)^4 (-1)^2 = 15 \times 16 = 240\).
The coefficient of \(x^3\) is given by \(\binom{6}{3} (2)^3 (-1)^3 = 20 \times 8 \times (-1) = -160\).
(ii) For \((3x+1)(2-x)^6\), we need the coefficient of \(x^3\).
Consider the terms that contribute to \(x^3\):
- From \(3x \times \text{coefficient of } x^2 \text{ in } (2-x)^6\): \(3 \times 240 = 720\).
- From \(1 \times \text{coefficient of } x^3 \text{ in } (2-x)^6\): \(-160\).
Thus, the coefficient of \(x^3\) is \(720 - 160 = 560\).
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