Write down the first three terms of the binomial expansion \((1 + ax)^{n}\) in ascending powers of x. If the coefficients of x and x\(^{2}\) are 2 and \(\fr...
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Write down the first three terms of the binomial expansion \((1 + ax)^{n}\) in ascending powers of x. If the coefficients of x and x\(^{2}\) are 2 and \(\frac{3}{2}\) respectively, find the values of a and n.
Answer Details
The binomial expansion of \((1+ax)^n\) in ascending powers of \(x\) is:
\[(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \dots\]
So the first three terms are \(1,\; nax,\; \dfrac{n(n-1)}{2}a^2x^2\).
Forming the equations. The coefficient of \(x\) is \(2\) and the coefficient of \(x^2\) is \(\tfrac{3}{2}\):
\[na = 2 \quad\text{(1)}, \qquad \frac{n(n-1)}{2}a^2 = \frac{3}{2} \quad\text{(2)}.\]
From (1), \(a = \dfrac{2}{n}\). Substituting into (2):
\[\frac{n(n-1)}{2}\cdot\frac{4}{n^2} = \frac{3}{2} \;\Rightarrow\; \frac{2(n-1)}{n} = \frac{3}{2}.\]
Cross-multiplying: \(4(n-1) = 3n \Rightarrow 4n-4 = 3n \Rightarrow n = 4\).
Then \(a = \dfrac{2}{n} = \dfrac{2}{4} = \dfrac{1}{2}\).
Answer: \(a = \tfrac{1}{2},\; n = 4\). (Check: with these values the terms are \(1 + 2x + \tfrac{3}{2}x^2\), as required.)
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