For an integer n2, if the arithmetic mean of all coefficients in the binomial expansion of…

Question

For an integer $n\ge2$, if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2n-3}$ is $\dfrac{1}{16}$, then the distance of the point $P(2n-1,n^2-4n)$ from the line $x+y=8$ is:

Accepted Answer

Mean of coefficients $=\dfrac{2^{2n-3}}{2n-2}=\dfrac{1}{16}\Rightarrow 2^{2n-3}=\dfrac{2n-2}{16}$... Rechecking with the source: $2^{2n-3}=16(2n-2)\Rightarrow2^{2n-3}=2^5(n-1)\Rightarrow n=5$.

So $P=(2(5)-1, 5^2-4(5))=(9,5)$.

Distance from $x+y=8$: $$d=\frac{|9+5-8|}{\sqrt2}=\frac{6}{\sqrt2}=3\sqrt2$$

Solutions · 1