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Question-156979

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Question Number 156979 by cortano last updated on 18/Oct/21
Commented by cortano last updated on 18/Oct/21
Σ_(k=1) ^n (−1)^k C_k ^( n) =0
$$\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \:{C}_{{k}} ^{\:{n}} \:=\mathrm{0}\: \\ $$
Commented by cortano last updated on 18/Oct/21
prove the equality
$${prove}\:{the}\:{equality} \\ $$
Answered by FongXD last updated on 18/Oct/21
Binomial Expansion: (a+b)^n =Σ_(r=0) ^n C(n,r)a^(n−r) b^r set a=1 and b=−1 we get: (1−1)^n =Σ_(r=0) ^n C(n,r)(−1)^r ⇔ 0=C(n,0)+C(n,1)(−1)^1 +C(n,2)(−1)^2 +C(n,3)(−1)^3 +...+C(n,n)(−1)^n therefore, 1−C(n,1)+C(n,2)−C(n,3)+...+(−1)^n C(n,n)=0
$$\mathrm{Binomial}\:\mathrm{Expansion}:\:\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{n}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}\left(\mathrm{n},\mathrm{r}\right)\mathrm{a}^{\mathrm{n}−\mathrm{r}} \mathrm{b}^{\mathrm{r}} \\ $$$$\mathrm{set}\:\mathrm{a}=\mathrm{1}\:\mathrm{and}\:\mathrm{b}=−\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{get}:\:\left(\mathrm{1}−\mathrm{1}\right)^{\mathrm{n}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}\left(\mathrm{n},\mathrm{r}\right)\left(−\mathrm{1}\right)^{\mathrm{r}} \\ $$$$\Leftrightarrow\:\mathrm{0}=\mathrm{C}\left(\mathrm{n},\mathrm{0}\right)+\mathrm{C}\left(\mathrm{n},\mathrm{1}\right)\left(−\mathrm{1}\right)^{\mathrm{1}} +\mathrm{C}\left(\mathrm{n},\mathrm{2}\right)\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{C}\left(\mathrm{n},\mathrm{3}\right)\left(−\mathrm{1}\right)^{\mathrm{3}} +…+\mathrm{C}\left(\mathrm{n},\mathrm{n}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\mathrm{therefore},\:\mathrm{1}−\mathrm{C}\left(\mathrm{n},\mathrm{1}\right)+\mathrm{C}\left(\mathrm{n},\mathrm{2}\right)−\mathrm{C}\left(\mathrm{n},\mathrm{3}\right)+…+\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{C}\left(\mathrm{n},\mathrm{n}\right)=\mathrm{0} \\ $$
Answered by puissant last updated on 18/Oct/21
f(x)=(1+x)^n =Σ_(k=0) ^n C_k ^n x^k 1^(n−k) ⇒ f(x)=(1+x)^n =Σ_(k=0) ^n C_k ^n x^k ; x=−1 ⇒ f(−1)=(1−1)^n =Σ_(k=0) ^n (−1)^k C_k ^n ⇒ Σ_(k=0) ^n (−1)^k C_k ^n = 0 ....
$${f}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {x}^{{k}} \mathrm{1}^{{n}−{k}} \\ $$$$\Rightarrow\:{f}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {x}^{{k}} \:;\:\:{x}=−\mathrm{1} \\ $$$$\Rightarrow\:{f}\left(−\mathrm{1}\right)=\left(\mathrm{1}−\mathrm{1}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{{n}} \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{k}} ^{{n}} =\:\mathrm{0}\:…. \\ $$

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