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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=0ij2019(j2019)(ij)
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=j=02019(i=0jCji)C2019ji=0jCji=i=0jCji1i1ji=2jx=j=02019C2019j2j12019j=32019

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