Question

When (a+(r/2))^5 is expanded, the coefficient of r^3 is 5/36. Find possible values of a.

Answer 1

Answer:

$a=\pm \dfrac{1}{3}$

Step-by-step explanation:

Use formula for binomial expansion:

$(a+b)^n=C_n^0a^nb^0+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+\dots+C_n^{n-1}a^1b^{n-1}+C_n^na^0b^n.$

Now

$\left(a+\dfrac{r}{2}\right)^5=C_5^0a^5\left(\dfrac{r}{2}\right)^0+C_5^1a^4\left(\dfrac{r}{2}\right)^1+C_5^2a^3\left(\dfrac{r}{2}\right)^2+C_5^3a^2\left(\dfrac{r}{2}\right)^3+C_5^4a^1\left(\dfrac{r}{2}\right)^4+C_5^5a^0\left(\dfrac{r}{2}\right)^5=\\ \\=a^5+5a^4\cdot \dfrac{r}{2}+10a^3\cdot \dfrac{r^2}{4}+10a^2\cdot \dfrac{r^3}{8}+5a\cdot \dfrac{r^4}{16}+\dfrac{r^5}{32}.$

The coefficient of $r^3$ is $\dfrac{10a^2}{8}.$ Since this coefficient is $\dfrac{5}{36},$ then

$\dfrac{10a^2}{8}=\dfrac{5}{36},\\ \\a^2=\dfrac{1}{9},\\ \\a=\pm \dfrac{1}{3}.$