Question

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

Answer 1

`a=\pm (1)/(3)`

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+(r)/(2)\right)^5=C_5^0a^5\left((r)/(2)\right)^0+C_5^1a^4\left((r)/(2)\right)^1+C_5^2a^3\left((r)/(2)\right)^2+C_5^3a^2\left((r)/(2)\right)^3+C_5^4a^1\left((r)/(2)\right)^4+C_5^5a^0\left((r)/(2)\right)^5=\\ \\=a^5+5a^4\cdot (r)/(2)+10a^3\cdot (r^2)/(4)+10a^2\cdot (r^3)/(8)+5a\cdot (r^4)/(16)+(r^5)/(32).`

The coefficient of
`r^3` is
`(10a^2)/(8).` Since this coefficient is
`(5)/(36),` then

`(10a^2)/(8)=(5)/(36),\\ \\a^2=(1)/(9),\\ \\a=\pm (1)/(3).`