Question
In the binomial expansion of \(\left(x+\frac{k}{x}\right)^7\), where \(k\) is a positive constant, the coefficients of \( x^3 \) and \(x\)are the same.
(i) Find the value of \(k\).
(ii) Using the value of \(k\) found in part (i), find the coefficient of \(x^7\) in the expansion of \(\left(1-5 x^2\right)\left(x+\frac{k}{x}\right)^7\).
Recognise that \(x\) is present in both terms within the brackets, this is a clue to use the general term $${T}_{r+1}={n \choose r} a^{n-r} b^r$$
$$\begin{aligned} & T_{r+1} \text { in }\left(x+\frac{k}{x}\right)^7 \\ & =\binom{7}{r}(x)^{7-r}\left(\frac{k}{x}\right)^r \\ & =\binom{7}{r}(x)^{7-r}(k)^r(x)^{-r} \\ & =\binom{7}{r} k^r(x)^{7-2 r}\end{aligned}$$
To be able to simplify the \({T}_{r+1}\) term correctly requires good grasp of indices
For term in \(x^3, 7-2 r=3\)
\(
r=2
\)
Term in \(x^3=\binom{7}{2} k^2(x)^3\)
\(
=21 k^2 x^3
\)
For term in \(x, 7-2 r=1\)
\(
r=3
\)
Term in \(x=\binom{7}{3} k^3(x)^1\) \(=35 k^3 x\)