# A proving question based on binomial theorem [closed]

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

## closed as off-topic by Saad, Paul Frost, Cesareo, user416281, NamasteDec 2 '18 at 15:33

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• What are the $C_k$ ? What are $a,b,c$? – darij grinberg Dec 2 '18 at 3:51
• $C_k$ is the binomial coefficient $\binom nk$ – Shubham Johri Dec 2 '18 at 5:23
• @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 $abcC_0$. – Shubham Johri Dec 2 '18 at 6:56

## 2 Answers

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.

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

$$(x-1)^n=C_0x^n-C_1x^{n-1}+C_2x^{n-2}...+(-1)^nC_n\\$$

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 $$x$$,

$$n(x-1)^{n-1}x^{a-n}+(a-n)(x-1)^nx^{a-n-1}=aC_0x^{a-1}-(a-1)C_1x^{a-2}+(a-2)C_2x^{a-3}...+(-1)^n(a-n)C_nx^{a-n-1}\\$$

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 $$x$$,

$$n(n-1)(x-1)^{n-2}x^{b-n+1}+n(b-n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}+n(a-n)(x-1)^{n-1}x^{b-n}=abC_0x^{b-1}-(a-1)(b-1)C_1x^{b-2}+(a-2)(b-2)C_2x^{b-3}...+(-1)^n(a-n)(b-n)C_nx^{b-n-1}\\$$

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

$$x^{c-b+1}[n(n-1)(x-1)^{n-2}x^{b-n+1}+n(a+b-2n+1)(x-1)^{n-1}x^{b-n}+(a-n)(b-n)(x-1)^nx^{b-n-1}]=n(n-1)(x-1)^{n-2}x^{c-n+2}+n(a+b-2n+1)(x-1)^{n-1}x^{c-n+1}+(a-n)(b-n)(x-1)^nx^{c-n}=abC_0x^c-(a-1)(b-1)C_1x^{c-1}+(a-2)(b-2)C_2x^{c-2}...+(-1)^n(a-n)(b-n)C_nx^{c-n}\\$$

Differentiate with respect to $$x$$,

$$n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(c-n+2)(x-1)^{n-2}x^{c-n+1}+n(n-1)(a+b-2n+1)(x-1)^{n-2}x^{c-n+1}+n(c-n+1)(a+b-2n+1)(x-1)^{n-1}x^{c-n}+n(a-n)(b-n)(x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=n(n-1)(n-2)(x-1)^{n-3}x^{c-n+2}+n(n-1)(a+b+c-3n+3)(x-1)^{n-2}x^{c-n+1}+n[(c-n+1)(a+b-2n+1)+(a-n)(b-n)](x-1)^{n-1}x^{c-n}+(a-n)(b-n)(c-n)(x-1)^nx^{c-n-1}=abcC_0x^{c-1}-(a-1)(b-1)(c-1)C_1x^{c-2}+(a-2)(b-2)(c-2)C_2x^{c-3}...+(-1)^n(a-n)(b-n)(c-n)C_nx^{c-n-1}\\$$

Set $$n>3, x=1$$,

$$0=abcC_0-(a-1)(b-1)(c-1)C_1+(a-2)(b-2)(c-2)C_2...+(-1)^n(a-n)(b-n)(c-n)C_n\\$$

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 $$abcC_0$$.