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Question Number 3001 by Filup last updated on 02/Dec/15

Can you evaluate: S=Σ_(i=k) ^n ^n C_i

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{evaluate}: \\ $$$${S}=\underset{{i}={k}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{i}} \\ $$

Commented by Rasheed Soomro last updated on 02/Dec/15

S=Σ_(i=k) ^n ^n C_i =^n C_k +^n C_(k+1) +^n C_(k+2) +...+^n C_n =((n!)/(k!(n−k)!))+((n!)/((k+1)!(n−k−1)!))+((n!)/((k+2)!(n−k−2)!))+...+((n!)/(n!(n−n)!))

$${S}=\underset{{i}={k}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{i}} \\ $$$$\:\:\:\:=^{{n}} {C}_{{k}} +^{{n}} {C}_{{k}+\mathrm{1}} +^{{n}} {C}_{{k}+\mathrm{2}} +...+^{{n}} {C}_{{n}} \\ $$$$=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}+\mathrm{1}\right)!\left({n}−{k}−\mathrm{1}\right)!}+\frac{{n}!}{\left({k}+\mathrm{2}\right)!\left({n}−{k}−\mathrm{2}\right)!}+...+\frac{{n}!}{{n}!\left({n}−{n}\right)!} \\ $$

Answered by Yozzi last updated on 02/Dec/15

S=Σ_(i=k) ^n ((n),(i) ) S=Σ_(i=0) ^n ((n),(i) )−Σ_(i=0) ^(k−1) ((n),(i) ) Now, (1+1)^n =Σ_(i=0) ^n ((n),(i) ) 1^(n−i) 1^i ∴ Σ_(i=0) ^n ((n),(i) )=2^n ∴ S=2^n −Σ_(i=0) ^(k−1) ((n),(i) ) k≤n⇒k−1≤n−1. There is then the possibility that Σ_(i=0) ^(k−1) ((n),(i) )=Σ_(i=0) ^(n−1) ((n),(i) )=Σ_(i=0) ^n ((n),(i) )− ((n),(n) )=2^n −1 or Σ_(i=0) ^(k−1) ((n),(i) )=Σ_(i=0) ^(n−2) ((n),(i) )=2^n − ((n),(n) )− ((n),((n−1)) )=2^n −1−n orΣ_(i=0) ^(k−1) ((n),(i) )=Σ_(i=0) ^(n−2) ((n),(i) )=2^n −1−n−((n(n−1))/(2!)) or.... down to when k=1.

$${S}=\underset{{i}={k}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}\:\: \\ $$$${S}=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}−\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix} \\ $$$${Now},\:\left(\mathrm{1}+\mathrm{1}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}\:\mathrm{1}^{{n}−{i}} \mathrm{1}^{{i}} \\ $$$$\therefore\:\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\mathrm{2}^{{n}} \: \\ $$$$\therefore\:{S}=\mathrm{2}^{{n}} −\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix} \\ $$$${k}\leqslant{n}\Rightarrow{k}−\mathrm{1}\leqslant{n}−\mathrm{1}. \\ $$$${There}\:{is}\:{then}\:{the}\:{possibility}\:{that}\: \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}−\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}=\mathrm{2}^{{n}} −\mathrm{1} \\ $$$${or}\:\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\mathrm{2}^{{n}} −\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}−\begin{pmatrix}{{n}}\\{{n}−\mathrm{1}}\end{pmatrix}=\mathrm{2}^{{n}} −\mathrm{1}−{n} \\ $$$${or}\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\begin{pmatrix}{{n}}\\{{i}}\end{pmatrix}=\mathrm{2}^{{n}} −\mathrm{1}−{n}−\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!} \\ $$$${or}....\:{down}\:{to}\:{when}\:{k}=\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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