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Question Number 55784 by gunawan last updated on 04/Mar/19

If n an odd natural number, then Σ_(r=0) ^n (((−1)^r )/(^n C_r )) equals

$$\mathrm{If}\:{n}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{natural}\:\mathrm{number},\:\mathrm{then} \\ $$$$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{r}} }{\:^{{n}} {C}_{{r}} \:}\:\mathrm{equals} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19

S=T_0 +T_1 +T_2 +...+T_n =(((−1)^0 )/((n!)/(0!(n−0)!)))+(((−1)^1 )/((n!)/(1!(n−1)!)))+(((−1)^2 )/((n!)/(2!(n−2)!)))+(((−1)^3 )/((n!)/(3!(n−3)!)))+...+(((−1)^n )/((n!)/(n!(n−n)!))) =((0!(n−0)!)/(n!))−((1!(n−1)!)/(n!))+((2!(n−2)!)/(n!))−((3!(n−3)!)/(n!))+..−((n!(n−n)!)/(n!)) now look T_0 =((0!(n−0)!)/(n!))=1 T_n =((−n!(n−n)!)/(n!))=−1 T_0 +T_n =0 T_1 =((−1!(n−1)!)/(n!))=((−1)/(n−1)) T_(n−1) =(((−1)^(n−1) ×(n−1)!(1)!)/(n!))=(1/(n−1)) T_1 +T_(n−1) =0 now let n=2k+1 terms →k=((n−1)/2) middle term is (k+1)_(th) term k+1→(((n−1)/2)+1)→((n+1)/2) k+1 th term→((n+1)/2) term→(((−1)^((n+1)/2) )/(nc_((n+1)/2) )) =(((((n+1)/2))!(n−((n+1)/2))!)/(n!))→(((((n+1)/2))!(((n−1)/2))!)/(n!)) now on addition all terms ancelled each other as explained above...only one term remain so S=(((((n+1)/2))!(((n−1)/2))!)/n)←this is the answer

$${S}={T}_{\mathrm{0}} +{T}_{\mathrm{1}} +{T}_{\mathrm{2}} +...+{T}_{{n}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{0}} }{\frac{{n}!}{\mathrm{0}!\left({n}−\mathrm{0}\right)!}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{1}} }{\frac{{n}!}{\mathrm{1}!\left({n}−\mathrm{1}\right)!}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} }{\frac{{n}!}{\mathrm{2}!\left({n}−\mathrm{2}\right)!}}+\frac{\left(−\mathrm{1}\right)^{\mathrm{3}} }{\frac{{n}!}{\mathrm{3}!\left({n}−\mathrm{3}\right)!}}+...+\frac{\left(−\mathrm{1}\right)^{{n}} }{\frac{{n}!}{{n}!\left({n}−{n}\right)!}} \\ $$$$ \\ $$$$=\frac{\mathrm{0}!\left({n}−\mathrm{0}\right)!}{{n}!}−\frac{\mathrm{1}!\left({n}−\mathrm{1}\right)!}{{n}!}+\frac{\mathrm{2}!\left({n}−\mathrm{2}\right)!}{{n}!}−\frac{\mathrm{3}!\left({n}−\mathrm{3}\right)!}{{n}!}+..−\frac{{n}!\left({n}−{n}\right)!}{{n}!} \\ $$$$ \\ $$$${now}\:{look}\:{T}_{\mathrm{0}} =\frac{\mathrm{0}!\left({n}−\mathrm{0}\right)!}{{n}!}=\mathrm{1} \\ $$$${T}_{{n}} =\frac{−{n}!\left({n}−{n}\right)!}{{n}!}=−\mathrm{1} \\ $$$${T}_{\mathrm{0}} +{T}_{{n}} =\mathrm{0} \\ $$$$ \\ $$$${T}_{\mathrm{1}} =\frac{−\mathrm{1}!\left({n}−\mathrm{1}\right)!}{{n}!}=\frac{−\mathrm{1}}{{n}−\mathrm{1}} \\ $$$${T}_{{n}−\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ×\left({n}−\mathrm{1}\right)!\left(\mathrm{1}\right)!}{{n}!}=\frac{\mathrm{1}}{{n}−\mathrm{1}} \\ $$$${T}_{\mathrm{1}} +{T}_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$$ \\ $$$${now} \\ $$$${let}\:{n}=\mathrm{2}{k}+\mathrm{1}\:{terms}\:\rightarrow{k}=\frac{{n}−\mathrm{1}}{\mathrm{2}} \\ $$$${middle}\:{term}\:{is}\:\left({k}+\mathrm{1}\right)_{{th}} \:{term} \\ $$$${k}+\mathrm{1}\rightarrow\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)\rightarrow\frac{{n}+\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${k}+\mathrm{1}\:{th}\:{term}\rightarrow\frac{{n}+\mathrm{1}}{\mathrm{2}}\:{term}\rightarrow\frac{\left(−\mathrm{1}\right)^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} }{{nc}_{\frac{{n}+\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)!\left({n}−\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)!}{{n}!}\rightarrow\frac{\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)!\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)!}{{n}!} \\ $$$$ \\ $$$${now}\:{on}\:{addition}\:{all}\:{terms}\:{ancelled}\:{each}\:{other} \\ $$$${as}\:{explained}\:{above}...{only}\:{one}\:{term}\:{remain} \\ $$$${so}\:\boldsymbol{{S}}=\frac{\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)!\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right)!}{{n}}\leftarrow{this}\:{is}\:{the}\:{answer} \\ $$$$ \\ $$$$ \\ $$

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