Expanding and Analyzing (1 - (2/a)x)(a+x)^5
The question is asking us to show that in the expansion of \((1 - \frac{2}{a}x)(a+x)^5\), where \(a\) is a non-zero constant, the coefficient of \(x^2\) is zero.
To find the coefficient of \(x^2\), we need to consider how terms which multiply together to give an \(x^2\) term could arise in the expansion. We can expand \((a+x)^5\) using the binomial theorem and then multiply the result by \((1 - \frac{2}{a}x)\). We're only interested in terms that give \(x^2\) after this multiplication, so we can ignore higher powers of \(x\).
The binomial expansion of \((a + x)^5\) is given by:
\((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5\)
We are interested in the terms up to \(x^2\) since multiplying by \(x^3\) or higher from the linear term outside will exceed \(x^2\).
When we multiply \((1 - \frac{2}{a}x)\) by the \(a^5\), \(5a^4x\), and \(10a^3x^2\) terms, we'll get coefficients for \(x^0\), \(x^1\), and \(x^2\), respectively. Terms resulting from \(10a^2x^3\) and beyond multiplied by \(1\) would be of a higher power than \(x^2\), and are therefore irrelevant. So we focus on the following:
\((1 - \frac{2}{a}x)(a^5 + 5a^4x + 10a^3x^2)\)
Expanding the relevant terms:
\(a^5(1 - \frac{2}{a}x) = a^5 - 2a^4x\)
\(5a^4x(1 - \frac{2}{a}x) = 5a^4x - 10a^3x^2\)
\(10a^3x^2(1 - \frac{2}{a}x) = 10a^3x^2 - 20a^2x^3\)
Now we can add up the coefficients of \(x^2\) from these products:
From \(a^5(1 - \frac{2}{a}x)\), we do not have an \(x^2\) term.
From \(5a^4x(1 - \frac{2}{a}x)\), the coefficient of \(x^2\) is \(-10a^3\).
From \(10a^3x^2(1 - \frac{2}{a}x)\), the coefficient of \(x^2\) is \(10a^3\).
If we sum the coefficients for \(x^2\) from these two terms, we get:
\(-10a^3 + 10a^3 = 0\)
Hence, the coefficient of \(x^2\) in the expansion of \((1 - \frac{2}{a}x)(a+x)^5\) is zero, as required to be shown.
