Product

srry my above post is wrong

ignore

I have another proof, w/o calculus

but i feel i≤j is ok...

can't we separate i<j and i=j

and when i= j
it is simply Σ(3/(2-a_i))^2 , which is = (Σ 3/(2-a_i))^2 - 2Σ(1≤i<j≤n)9(2-a_i)(2-a_j) ??

@ HS Sir, Oh yes sir, It's i < j

\sum_{1\leq i<j\leq n}^{{}}\frac{1}{(2-a_i)(2-a_j)}

= \left [ x^{n-2} \right ]\prod_{i=1}^{n}\left ( x-\frac{1}{(2-a_i)} \right )

= \left [ x^{n-2} \right ]\; \; x^n\prod_{i=1}^{n}\left ( 1-\frac{1}{x(2-a_i)}\right )

= \left [ x^{n-2} \right ]\; \; \frac{x^n}{\prod_{i=1}^{n}(2-a_i)}\prod_{i=1}^{n}\left ( 2-a_i - \frac{1}{x} \right )

= \left [ x^{n-2} \right ]\; \; \frac{x^n}{2^n-1}\prod_{i=1}^{n}\left ( \left ( 2-\frac 1 x \right ) -a_i\right )

=\left [ x^{n-2} \right ]\; \; \frac{x^n}{2^n-1}\left ( \left ( 2-\frac 1 x \right )^n-1 \right )

= \left [ x^{n-2} \right ]\; \; \frac{\left ( 2x-1 \right )^n - x^n}{2^n-1}

= \frac{(2^{n-2})\cdot\binom{n}{2}}{2^n-1}

If r_1, r_2,..,r_n are the n roots of a nth degree polynomial f(x), then The expression

\sum_{1\le i<j \le n} \frac{1}{(x-r_i)(x-r_j)}

is equivalent to (when x is not one of the roots)

\frac{f"(x)}{2 f(x)}

You can easily prove this in atleast two ways and I suggest you try it out yourself.

In the given case, f(x)=x^n-1.

Hence

\sum_{1\le i<j \le n} \frac{1}{(x-r_i)(x-r_j)} = n(n-1)\frac{x^{n-2}}{2(x^n-1)}

Put x=2 and finish

One way to prove it:

We assume WLOG that the leading coefficient is 1.

So f(x) = (x-r_1)(x-r_2)...(x-r_n)

Now if you differentiate, you will get

\frac{df(x)}{dx} = (x-r_2)...(x-r_n)+(x-r_1)(x-r_3)...(x-r_n)+...+(x-r_1)(x-r_2)...(x-r_{n-1})

I guess you already know that if you now divide by f(x), the expression becomes

\frac{1}{x-r_1}+\frac{1}{x-r_2}+...+\frac{1}{x-r_n}

Differentiate f'(x) again and divide by f(x) and you will get \frac{f"(x)}{f(x)}=2 \sum_{1\lei<j\len} \frac{1}{(x-r_i)(x-r_j)}

Shukriya Sir!