Write down and simplify, in ascending powers of \(x\), the first four terms of the expansion of \((1-x)^{12}\).[2]
Hence find the value of \(p\) given that the coefficient of \(x^3\) in the expansion of \((2x^2+17x+p)(1-x)^{12}\) is \(6598\).[3]
In the binomial expansion of \(\left(x+\dfrac{k}{x}\right)^5\), where \(k\) is a positive constant, the coefficients of \(x^3\) and \(x\) are the same. Find the value of \(k\).[4]
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Answer:(a)(i) \(1-12x+66x^2-220x^3\); (ii) \(p=-25\). (b) \(k=\dfrac12\).