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let-u-n-k-1-n-1-k-n-k-calculate-n-1-u-n-




Question Number 45957 by maxmathsup by imad last updated on 19/Oct/18
let u_n =Σ_(k=1) ^n  (1/(k!(n−k)!))  calculate Σ_(n=1) ^∞ u_n
$${let}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}!\left({n}−{k}\right)!}\:\:{calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} {u}_{{n}} \\ $$
Commented by maxmathsup by imad last updated on 20/Oct/18
we have u_n =(1/(n!)) Σ_(k=1) ^n    ((n!)/(k!(n−k)!)) =(1/(n!)) Σ_(k=1) ^n  C_n ^k   =(1/(n!)) (Σ_(k=0) ^n  C_n ^k −1) =(1/(n!)){ 2^n −1} =(2^n /(n!)) −(1/(n!)) ⇒  Σ_(n=1) ^∞   u_n = Σ_(n=1) ^∞  (2^n /(n!)) −Σ_(n=1) ^∞  (1/(n!)) = Σ_(n=0) ^∞  (2^n /(n!)) −1 −Σ_(n=0) ^∞  (1/(n!)) +1  =e^2  −e .
$${we}\:{have}\:{u}_{{n}} =\frac{\mathrm{1}}{{n}!}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:=\frac{\mathrm{1}}{{n}!}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \\ $$$$=\frac{\mathrm{1}}{{n}!}\:\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} −\mathrm{1}\right)\:=\frac{\mathrm{1}}{{n}!}\left\{\:\mathrm{2}^{{n}} −\mathrm{1}\right\}\:=\frac{\mathrm{2}^{{n}} }{{n}!}\:−\frac{\mathrm{1}}{{n}!}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{u}_{{n}} =\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}^{{n}} }{{n}!}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{{n}} }{{n}!}\:−\mathrm{1}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:+\mathrm{1} \\ $$$$={e}^{\mathrm{2}} \:−{e}\:. \\ $$
Answered by Smail last updated on 19/Oct/18
u_n =Σ_(k=1) ^n (1/(k!(n−k)!))=Σ_(k=1) ^n ((n!)/(k!(n−k)!))×(1/(n!))  =(1/(n!))Σ_(k=1) ^n ((n!)/(k!(n−k)!))=(1/(n!))(Σ_(k=0) ^n ((n!)/(k!(n−k)!))−1)  let′s take p(x)=(1+x)^n =Σ_(k=0) ^n ((n!)/(k!(n−k)!))x^k   So p(1)=2^n =Σ_(k=0) ^n ((n!)/(k!(n−k)!))  u_n =(1/(n!))(2^n −1)  Σ_(n=1) ^∞ u_n =Σ_(n=1) ^∞ (2^n /(n!))−Σ_(n=1) ^∞ (1/(n!))  e^x =Σ_(n=0) ^∞ (x^n /(n!))  e^1 =Σ_(n=0) ^∞ (1/(n!))  e^2 =Σ_(n=0) ^∞ (2^n /(n!))  So Σ_(n=1) ^∞ u_n =Σ_(n=0) ^∞ (2^n /(n!))−1−(Σ_(n=0) ^∞ (1/(n!))−1)  =e^2 −e
$${u}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}!\left({n}−{k}\right)!}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}!}{{k}!\left({n}−{k}\right)!}×\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{\mathrm{1}}{{n}!}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}!}{{k}!\left({n}−{k}\right)!}=\frac{\mathrm{1}}{{n}!}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{{k}!\left({n}−{k}\right)!}−\mathrm{1}\right) \\ $$$${let}'{s}\:{take}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{{k}!\left({n}−{k}\right)!}{x}^{{k}} \\ $$$${So}\:{p}\left(\mathrm{1}\right)=\mathrm{2}^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$$${u}_{{n}} =\frac{\mathrm{1}}{{n}!}\left(\mathrm{2}^{{n}} −\mathrm{1}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{u}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{n}} }{{n}!}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!} \\ $$$${e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!} \\ $$$${e}^{\mathrm{1}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!} \\ $$$${e}^{\mathrm{2}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{n}} }{{n}!} \\ $$$${So}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{u}_{{n}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{{n}} }{{n}!}−\mathrm{1}−\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}−\mathrm{1}\right) \\ $$$$={e}^{\mathrm{2}} −{e} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 20/Oct/18
correct answer thanks...
$${correct}\:{answer}\:{thanks}… \\ $$

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