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The binomial expansion of \((a+bx)^{n}\) begins \(16-8x\dots\) If \(a,b,n\) are all integers, find the possible sets of values for \(a,b\) and \(n\).

If \(a, b\) are integers, but \(n\) is not, show that \(a=2^{k}\) where \(k≥3\) and find expressions for \(b\) and \(n\) in term of \(k\).


So \(a^{n}=16\) and \(\frac{a^{n}nbx}a=-8x\) which gives \(\frac{16nbx}a=-8\) so \(\frac{nb}a=-\frac 1 2\) and \(b=-\frac a {2n}\)

If \(a\) and \(n\) are both integers, either \(a=16\) or \(a=4\) or \(a=-4\) or \(a=2\)

If \(a=16, n=1, b=-\frac{16}2=-8\) which gives \((16-8x)^{1}=16-8x\)

If \(a=\pm4, n=2, b=\mp\frac4{2\times2}=\mp1\) which gives \((4-x)^{2}\) or \((-4+x)^{2}\)

If \(a=2, n=4, b=-\frac2{2\times8}\) which is not an integer so it does not work.

If \(n\) is negative than \(a^{n}=16\) gives a value of \(a\) between \(0\) and \(1\), so not possible.

If \(a\) is an integer and \(a^{n}=16\) then \(=(2^{4})^{1/n}=2\frac4 n\) is an integer. So \(n=\frac 4 k\)

This gives \(b=-\frac a{2n}=-\frac{2^{k}}{2\times\frac4 k}=-2^{k-3}k\) which is an integer for \(k≥3\)

(We have already seen that \(a=4, 16(k=2,4)\) give integer values of \(n\) and \(b\))

Alternative approach – use a spreadsheet to investigate using a list of integers for \(a\) with columns for \(n=\frac{\log 16}{\log a}\) and \(b=-\frac a{2n}\). It becomes clear that the powers of \(2\) give integers for \(b\) and rational values for \(n\). Then create a list of the powers of \(2\) and their corresponding values of \(n\) and \(b\) for further insight into the problem.

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