Saphia1123

I think it's somewhere around 284. Does it say it is a whole number? Or are they looking for a radical?

I would check this link out: https://math.stackexchange.com/questions/2460311/how-many-6-digit-numbers-can-be-formed-in-which-the-sum-if-its-digits-is-divisib

If we draw a parallel line to $AB$ and mark the intersection point $E$, we create an equilateral $\triangle ADE$. This means we can derive the following information: $$a_1 = AD, DE, AE$$ $$\triangle CDE ~ \triangle ABC*$$ $$\frac{DE}{AB}=\frac{CE}{AC}**$$ *by AA similarity ($120^{\circ}$ and $\angle C$) **by rules for similar triangles Notice in $\frac{DE}{AB}=\frac{CE}{AC}$ we can substitute $DE, AB, CE,$ and $AC$ with values of $a_1, b_1,$ and $c_1$. $$DE = a_1$$ $$AB = c_1$$ $$CE = b_1 - a_1$$ $$AC = b_1$$ Substituting those values, we can say: $$\frac{a}{c}=\frac{b-a}{b}$$ Multiplying both sides by $bc$ to eliminate the fraction, we get: $$ab = bc -ac$$ Add $ac$ to both sides: $$ab + ac = bc$$ Factor $a$ from the left-hand side: $$a(b+c)=bc$$ Divide both sides by $(b+c)$ and we get: $$a=\frac{bc}{b+c}$$ We can take the reciprocal of both sides to get: $$\frac{1}{a} = \frac{b+c}{bc}$$ Finally, we get: $$\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$$

The f(n) = 1/(n+1) is the simplified form of the function.

(2n)/n - (2n)/(n + 1) = f(n) (2n)/n

Simplifying:

(2n)/n - (2n)/(n + 1) = 2f(n)

(2n)/2n - (2n)/2(n + 1) = 2f(n)/2

1 - n/(n+1) = f(n)

(n+1 - n)/(n+1) = f(n)

1/(n+1) = f(n)

Ah ok. So then to prove the equality, I use part of the Cauchy-Schwarz to show it would be a degenerate triangle. And then the inequality part of Cauchy-Schwarz to prove it would apply for any positive values of a, b, and c because it would make a triangle?

Hey thank you. Any way to apply the law of cosines? I noticed that the terms of the inequality kinda looks like the law of cosines if you write it like this: a^2-2ab+b^2=cos C. That's difficult because there is a 2ab, and I don't know how to get rid of it. I think maybe involving trying to find the area of a triangle with 1/2 * ab sin angle C. I think I figured out the equality part because it would be a degenerate triangle. But how do I prove the inequality? I know that part is related to the triangle inequality theorem.

Hey Thank you for the response! Yea I was thinking along the lines of that to help cancel the square root, but I just am not sure how to go from there. I labeled a triangle with c= sqrt(a^2 - ab +b^2) b= sqrt(a^2 - ac +c^2) a= sqrt(b^2 + bc +c^2) Then, I tried putting values into the law of cosines and got: c^2= (b^2 + bc +c^2) + (a^2 - ac +c^2) -2(sqrt(b^2 + bc +c^2)*sqrt(a^2 - ac +c^2)) I don't know how this proves anything. I used the triangle inequality theorem to substitute the terms as a, b, and c in triangle ABC with the lower case letters representing the opposite angles.