Question - Expanding and Analyzing (1 - (2/a)x)(a+x)^5
Solution:
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.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com