For example, find, in ascending powers of $x$, the first three terms in the expansion of $(2+5x)^7$.
So, $(2+5x)^7=2^7+\binom{7}{1}(2^6)(5x)+\binom{7}{2}(2^5)(5x)^2$.
I've no problem to solve this kind of problem.
But now another question wants me to find, in descending powers of x, the first four terms in the expansion of $(2x+\frac{1}{3x})^6$.
How?
$\endgroup$1 Answer
$\begingroup$Exactly the same way, you can just do this with the binomial theorem.
\begin{align*} \left(2x+\frac{1}{3x}\right)^6 &= \sum_{k=0}^6 {6 \choose k}(2x)^{6-k}\left(\frac{1}{3x}\right)^k \\ &= 2^6x^6+2^5\frac{1}{3}{6 \choose 1} x^4 + 2^4\frac{1}{3^2}{6 \choose 2} x^2+ 2^3\frac{1}{3^3}{6 \choose 3} + \cdots \\ &= 64x^6+ 64x^4+\frac{80}{3} x^2 + \frac{160}{27}+\cdots \end{align*}
The terms $x^5$, $x^3$ and $x$ have coefficient 0.
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