$$C_0-C1(a-1)(b-1)(c-1)_+C_2(a-2)(b-2)(c-2)+.... (-1)^nC_n(a-n)(b-n)(c-n) $$=0 I tried to solve this problem by using multinomial theorem but was not able to proceed further please help me out.


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  • $\begingroup$ What are the $C_k$ ? What are $a,b,c$? $\endgroup$ – darij grinberg Dec 2 '18 at 3:51
  • $\begingroup$ $C_k$ is the binomial coefficient $\binom nk$ $\endgroup$ – Shubham Johri Dec 2 '18 at 5:23
  • $\begingroup$ @priyanka kumari Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4,a=b=c=1/2$, and it seems it should be $abcC_0$. $\endgroup$ – Shubham Johri Dec 2 '18 at 6:56

From \begin{align*} (1-x)^n = \sum (-1)^i \binom{n}{i} x^i \end{align*} differentiating, we obtain \begin{align*} n(1-x)^{n-1} = \sum (-1)^ii \binom{n}{i} x^{i-1} \end{align*} Multiply by $x$ and differentiate: \begin{align*} n(1-x)^{n-1}x &= \sum (-1)^ii \binom{n}{i} x^{i}\\ n(n-1)(1-x)^{n-2}x + n(1-x)^{n-1} = \sum (-1)^i i^2 \binom{n}{i} x^{i-1} \end{align*} Use the same technique for obtaining an expression for $\sum (-1)^i i^3 x^i \binom{n}{i}$.\ The given expression can be expanded as \begin{align*} \sum (-1)^i \binom{n}{i} [abc - i(a+b+c) + i^2(ab+bc+ca) -i^3] \end{align*} Put $x=1$ and see that all the individual expressions evaluate to 0.

  • $\begingroup$ Not able to understand please tell some other method. $\endgroup$ – priyanka kumari Dec 2 '18 at 5:31


Multiply by $x^{a-n}$,

$\implies (x-1)^nx^{a-n}=C_0x^a-C_1x^{a-1}+C_2x^{a-2}...+(-1)^nC_nx^{a-n}\\$

Differentiate with respect to $x$,


Multiply with $x^{b-a+1}$,

$\implies x^{b-a+1}[n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}]=n(x-1)^{n-1}x^{b-n+1}+(a-n)(x-1)^nx^{b-n}=aC_0x^b-(a-1)C_1x^{b-1}+(a-2)C_2x^{b-2}...+(-1)^n(a-n)C_nx^{b-n}\\$

Differentiate with respect to $x$,


Multiply with $x^{c-b+1}$,


Differentiate with respect to $x$,


Set $n>3, x=1$,


Are you sure it's $C_0$ and not $abcC_0$? I checked for the case $n=4, a=b=c=1/2$, and it seems it should be $abcC_0$.


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