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x-0-i-j-2019-j-2019-i-j-




Question Number 60725 by naka3546 last updated on 25/May/19
x  =  Σ_(0≤i≤j≤2019)  ( _(    j)^(2019)  )(  _i^j   )
$${x}\:\:=\:\:\underset{\mathrm{0}\leqslant{i}\leqslant{j}\leqslant\mathrm{2019}} {\sum}\:\left(\:_{\:\:\:\:{j}} ^{\mathrm{2019}} \:\right)\left(\:\:_{{i}} ^{{j}} \:\:\right)\: \\ $$
Commented by naka3546 last updated on 25/May/19
x  =  ?
$${x}\:\:=\:\:? \\ $$
Commented by Mr X pcx last updated on 25/May/19
x= Σ_(j=0) ^(2019)  (Σ_(i=0) ^j  C_j ^i )C_(2019) ^j   Σ_(i=0) ^j  C_j ^i  =Σ_(i=0) ^j  C_j ^i  1^i  1^(j−i)  =2^j  ⇒  x =Σ_(j=0) ^(2019)  C_(2019) ^j  2^j  1^(2019−j)   =3^(2019)
$${x}=\:\sum_{{j}=\mathrm{0}} ^{\mathrm{2019}} \:\left(\sum_{{i}=\mathrm{0}} ^{{j}} \:{C}_{{j}} ^{{i}} \right){C}_{\mathrm{2019}} ^{{j}} \\ $$$$\sum_{{i}=\mathrm{0}} ^{{j}} \:{C}_{{j}} ^{{i}} \:=\sum_{{i}=\mathrm{0}} ^{{j}} \:{C}_{{j}} ^{{i}} \:\mathrm{1}^{{i}} \:\mathrm{1}^{{j}−{i}} \:=\mathrm{2}^{{j}} \:\Rightarrow \\ $$$${x}\:=\sum_{{j}=\mathrm{0}} ^{\mathrm{2019}} \:{C}_{\mathrm{2019}} ^{{j}} \:\mathrm{2}^{{j}} \:\mathrm{1}^{\mathrm{2019}−{j}} \\ $$$$=\mathrm{3}^{\mathrm{2019}} \\ $$

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