# How to prove $\sum_{i=1}^n\frac{(1-a_i)^n}{a_i\prod_{j\neq i}(a_j-a_i)}=\frac{1}{a_1\cdots a_n}-1$?

Prove that $$\displaystyle\sum_{i=1}^n\frac{(1-a_i)^n}{a_i\prod_{j\neq i}(a_j-a_i)}=\frac{1}{a_1\cdots a_n}-1$$ for distinct $$a_1,\cdots,a_n\in\mathbb{R}\backslash\{0\}$$.

I noticed that it is equivalent to $$\displaystyle\sum_{i=1}^nx_i^n\prod_{j\neq i}\frac{1-x_j}{x_i-x_j}=1-\prod_{k=1}(1-x_k)$$ where $$x_i=1-a_i$$, then the LHS can be regarded as Lagrange interpolation formula, the value at $$x=1$$ of the polynomial of degree $$n-1$$ going through $$(x_1,x_1^n),\cdots,(x_n,x_n^n)$$.

But I don't know if it is useful for the proof.

• You can edit a question instead of reposting it.
– Aig
Commented Mar 20 at 17:33
• Oh I see, thank you! Commented Mar 21 at 2:41

You are almost done using your approach. You have an $$n-1$$ degree polynomial $$p(x)$$ such that $$p(a_i) = a_i^n$$. The $$a_i$$ are the roots of $$x^n - p(x)$$ and since this is monic, it follows that $$x^n - p(x) = (x-a_1)...(x-a_n)$$. Then, plugging in 1 gives $$p(1) = 1 - (1-a_1)...(1-a_n)$$ which is what you wanted.

• Got it, thank you very much! Commented Mar 21 at 2:38

Another approach, using partial fractions

Solve by partial fractions:

$$\frac{(x-1)^n}{(x-a_1)(x-a_2)\cdots (x-a_n)}=1+\sum_{i=1}^n \frac{b_i}{x-a_i}\tag1$$

Multiplying by $$x-a_i$$ and evaluating both sides at $$x=a_i,$$ you get:

$$b_i=\frac{(a_i-1)^n}{\prod_{j\neq i} (a_i-a_j)}=-\frac{(1-a_i)^n}{\prod_{j\neq i}(a_j-a_i)}$$

Evaluating $$(1)$$ at $$x=0,$$ you get:

$$\frac{1}{a_1\dots a_n}=1+\sum_{i=1}^n \frac{(1-a_i)^n}{a_i\prod_{j\neq i}(a_j-a_i)}$$

More generally, if $$p(x)$$ is a polynomial of degree less than or equal to $$n$$ with coefficient $$c$$ at $$x^n,$$ we get:

$$\frac{(-1)^np(0)}{a_1\dots a_n}=c-\sum_i\frac{p(a_i)}{a_i\prod_{j\neq i}(a_i-a_j)}$$

If $$p(x)=(x-k)^n,$$ then you get:

$$\frac{k^n}{a_1\dots a_n}-1=\sum_{i=1}^n\frac{(k-a_i)^n}{a_i\prod_j(a_j-a_i)}$$

If $$p(x)=(x-k)^m$$ for $$m then $$c=0$$ and:

$$\frac{k^m}{\prod a_i}=\sum_{i=1}^n \frac{(k-a_i)^m}{a_i\prod(a_j-a_i)}$$

If $$1\leq m\leq n$$ then you get, by subtracting $$k^m/\prod a_i$$ from $$k\cdot k^{m-1}/\prod a_i$$

$$\sum_{i=1}^n\frac{(k-a_i)^{m-1}}{\prod_j(a_j-a_i)}=\delta_{mn}=\begin{cases}1&m=n\\0&m\neq n\end{cases}.$$

• Got it, thank you very much! Commented Mar 21 at 2:39
• My answer isn't really that different from the Lagrange interpolation view - there is a strong relationship between partial fractions and Legrange interpolation of polynomials. I just think partial fractions first when I see such equalities. Commented Mar 21 at 14:18