Not Newton's Sum

Please upvote if you like it

---

Let $\alpha=a,\beta=b,\gamma=c$

Now as we know that the individual roots $a,b,c$ are making a geometric progression i.e we have to find

$a^0+a^1+a^2....a^n+b^0+b^1+b^2....b^n+c^0+c^1+c^2....c^n$

which will give us

$\dfrac{a^0}{1-a}+\dfrac{b^0}{1-b}+\dfrac{c^0}{1-a}$

$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}............(1)$

By vieta we have

$a+b+c=\dfrac{287}{245} ,ab+bc+ca=\dfrac{99}{245} ,abc=\dfrac{9}{245}$

on simplifying $..........(1)$ we get

$\dfrac{3-2(a+b+c)+ab+bc+ca}{1-(a+b+c)+ab+bc+ca-abc}$

Now plugging

$\dfrac{3-2(\dfrac{287}{245})+\dfrac{99}{245}}{1-(\dfrac{287}{245})+\dfrac{99}{245}-\dfrac{9}{245}}$

which gives us $\dfrac{65}{12}$

here $m=65,n=12$

$\therefore n+1=13$

$\therefore \dfrac{m}{n+1}=\dfrac{65}{13}=\huge\boxed{5}$

First simplify the infinite summation to $\frac {1}{1-\alpha}$ + $\frac {1}{1-\beta}$ + $\frac {1}{1-\gamma}$ . The cubic equation can also be written as $245x^3-287x^2+99x-9=245(x-\alpha)(x-\beta)(x-\gamma)$ .

Simplify $\frac {1}{1-\alpha}$ + $\frac {1}{1-\beta}$ + $\frac {1}{1-\gamma}$ to $\frac {\alpha\gamma+\beta\gamma+\alpha\beta -2(\alpha+\beta+\gamma)+3}{(1-\alpha)(1-\beta)(1-\gamma)}$ .

For the denominator plug in $x=1$ . So $(1-\alpha)(1-\beta)(1-\gamma)= \frac {48}{245}$ .

Use vieta's formula to find out that $\beta\gamma+\alpha\gamma+ \alpha\beta= \frac {99}{245}$ , and that $\alpha+\beta+\gamma=\frac{287}{245}$ .

Then plug in those values for the numerator $\frac{99}{245}- 2(\frac{287}{245})+3 = \frac{52}{49}$ .

Then simplify $\frac {\frac{52}{49}}{\frac{48}{245}}=\frac{65}{12}$ . So $m=65$ and $n=12$ . so $\frac {m}{n+1}= \frac{65}{13}$ = $\boxed{5}$ .

This is not a genuine cubic equation.

alpha = 1/ 7

beta = 3/ 5

gamma = 3/ 7

1/ (1 - 1/ 7) + 1/ (1 - 3/ 5) + 1/ (1 - 3/ 7) = 7/ 6+ 5/ 2 +7/ 4 = 65/ 12

65/ (12 + 1) = 5

I did it with Newton Sum method and cheated with a spreadsheet:

f(x)=(x-a)(x-b)(x-c) Take log and differentiate f'(x)/f(x) =(x-a)^-1+(x-b)^-1+(x-c)^-1 =1/x(1+a/x+a^2/x^2 ...)+1/x(1+b/x+b^2/x^2 ...)+1/x(1+c/x+c^2/x^2 ..) Now just put x=1 so m/n = f'(1)/f(1).=65/12 and m/n+1 =5 This solution has a advantage as if roots are more than 3 say n then also ans. will be f'(1)/f(1) you don't need deal with them individually making equations and solving them . This give solution to whole class of such problems.