${\displaystyle {\begin{aligned}&{\dot {p))=-{\frac {\partial \mathrm {H} }{\partial q))\\&{\dot {q))=+{\frac {\partial \mathrm {H} }{\partial p))\end{aligned))}$

${\displaystyle {\frac {1}{1+x))=(1+x)^{-1}=1-x+x^{2}-x^{3}+...=\sum _{k=0}^{\infty }(-1)^{k}x^{k))$

${\displaystyle (-1)(1+x)^{-2}=-1+2x-3x^{2}...=\sum _{k=1}^{\infty }(-1)^{k}kx^{k-1))$

${\displaystyle (-1)(-2)(1+x)^{-3}=\sum _{k=2}^{\infty }(-1)^{k}k(k-1)x^{k-2))$

${\displaystyle (-1)(-2)(-3)(1+x)^{-4}=\sum _{k=3}^{\infty }(-1)^{k}k(k-1)(k-2)x^{k-3}=\sum _{r=0}^{\infty }(-1)^{r+3}(r+3)(r+2)(r+1)x^{r))$

${\displaystyle (-1)^{n-1}(n-1)!(1+x)^{-n}=\sum _{l=0}^{\infty }(-1)^{r+n-1}{\frac {(r+n-1)!}{r!))x^{r))$

${\displaystyle (1+x)^{-n}=\sum _{l=0}^{\infty }(-1)^{r}{\frac {(r+n-1)!}{r!(n-1)!))x^{r))$

${\displaystyle {\binom {-n}{r))=(-1)^{r}{\frac {(r+n-1)!}{r!(n-1)!))=(-1)^{r}{\binom {r+n-1}{r))}$

${\displaystyle {\binom {-n-1}{r))=(-1)^{r}{\binom {r+n}{r))}$

${\displaystyle (-1)^{r}{\binom {-n-1}{r))={\binom {r+n}{r))}$

${\displaystyle (-1)^{r}{\binom {-n+r-1}{r))={\binom {n}{r))}$

${\displaystyle {\binom {9}{5))-{\binom {9}{6))+{\binom {9}{7))-{\binom {9}{8))+{\binom {9}{9))=126-84+36-9+1=256=70={\binom {8}{4))}$

${\displaystyle \sum _{k=0}^{n}(-1)^{k}{\binom {2n+1}{n+k+1))=\sum _{r=n}^{0}(-1)^{n-r}{\binom {2n+1}{2n-r+1))=\sum _{r=n}^{0}(-1)^{n-r}{\binom {2n+1}{r))=}$ ${\displaystyle \sum _{r=0}^{n}(-1)^{n-r}{\binom {2n+1}{r))=(-1)^{n}\sum _{r=0}^{n}(-1)^{r}{\binom {2n+1}{r))=}$

This 100

${\displaystyle \sum _{i=r}^{n}{\binom {n-i+1}{1)){\binom {i}{r))=\sum _{i=r}^{n}(n-i+1){\binom {i}{r))=\sum _{i=r}^{n}\sum _{j=i}^{n}{\binom {i}{r))=\sum _{j=r}^{n}\sum _{i=r}^{j}{\binom {i}{r))=\sum _{j=r}^{n}{\binom {j+1}{r+1))={\binom {n+2}{r+2))}$

therefore ${\displaystyle \sum _{i=0}^{n}{\binom {n-i+1}{1)){\binom {i}{0))={\binom {n+2}{2))}$

Then

${\displaystyle \sum _{i=r}^{n}{\binom {n-i+2}{2)){\binom {i}{r))=\sum _{i=r}^{n}\sum _{j=0}^{n}{\binom {n-i-j+1}{1)){\binom {j}{0)){\binom {i}{r))}$

${\displaystyle =\sum _{i=r}^{n}\sum _{j=0}^{n}{\binom {n-i-j+1}{1)){\binom {i}{r))}$ ${\displaystyle =\sum _{j=0}^{n}\sum _{i=r}^{n}{\binom {n-i-j+1}{1)){\binom {i}{r))=\sum _{j=0}^{n}{\binom {n-j+2}{r+2))=\sum _{j=2}^{n+2}{\binom {j}{r+2))={\binom {n+3}{r+3))}$

${\displaystyle \sum _{i=r}^{n}{\binom {n-i+k-1}{k-1)){\binom {i}{r))={\binom {n+k}{r+k))}$

${\displaystyle {\mathbf {\hat {r)) }={\frac {\mathbf {r} }{r))={\frac {x\mathbf {i} +y\mathbf {j} +z\mathbf {k} }{r))=\mathbf {i} \sin \theta \cos \phi +\mathbf {j} \sin \theta \sin \phi +\mathbf {k} \cos \theta }$

${\displaystyle \mathbf {\boldsymbol {\hat {\phi ))} ={\hat {\mathbf {\phi } ))={\frac {\mathbf {k} \times {\hat {\mathbf {r} ))}{\sin \theta ))=-\mathbf {i} \sin \phi +\mathbf {j} \cos \phi }$

${\displaystyle \mathbf {\boldsymbol {\hat {\theta ))} =\mathbf {\boldsymbol {\hat {\phi ))} \times \mathbf {\boldsymbol {\hat {r))} =\mathbf {i} \cos \theta \cos \phi +\mathbf {j} \cos \theta \sin \phi -\mathbf {k} \sin \theta }$

Wee see,

${\displaystyle {\frac {\partial \mathbf {\hat {r)) }{\partial r))=0}$

${\displaystyle {\frac {\partial \mathbf {\hat {r)) }{\partial \theta ))=\mathbf {i} \cos \theta \cos \phi +\mathbf {j} \cos \theta \sin \phi -\mathbf {k} \sin \theta =\mathbf {\boldsymbol {\hat {\theta ))} }$

${\displaystyle {\frac {\partial \mathbf {\hat {r)) }{\partial \phi ))=-\mathbf {i} \sin \theta \sin \phi +\mathbf {j} \sin \theta \cos \phi =(-\mathbf {i} \sin \phi +\mathbf {j} \cos \phi )\sin \theta =\mathbf {\boldsymbol {\hat {\phi ))} \sin \theta }$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\phi ))} }{\partial r))=0}$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\phi ))} }{\partial \theta ))=0}$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\phi ))} }{\partial \phi ))=-\mathbf {i} \cos \phi -\mathbf {j} \sin \phi =-(\mathbf {\hat {r)) \sin \theta +\mathbf {\boldsymbol {\hat {\theta ))} \cos \theta )}$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\theta ))} }{\partial r))=0}$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\theta ))} }{\partial \theta ))=-\mathbf {i} \sin \theta \cos \phi -\mathbf {j} \sin \theta \sin \phi -\mathbf {k} \cos \theta =-\mathbf {\hat {r)) }$

${\displaystyle {\frac {\partial \mathbf {\boldsymbol {\hat {\theta ))} }{\partial \phi ))=-\mathbf {i} \cos \theta \sin \phi +\mathbf {j} \cos \theta \cos \phi ={\boldsymbol {\hat {\phi ))}\cos \theta }$

${\displaystyle d\mathbf {r} =d(r\mathbf {\hat {r)) )=\mathbf {\hat {r)) dr+rd\mathbf {\hat {r)) =\mathbf {\hat {r)) dr+r\left({\frac {\partial \mathbf {\hat {r)) }{\partial r))dr+{\frac {\partial \mathbf {\hat {r)) }{\partial \theta ))d\theta +{\frac {\partial \mathbf {\hat {r)) }{\partial \phi ))d\phi \right)}$

${\displaystyle =\mathbf {\hat {r)) dr+\mathbf {\boldsymbol {\hat {\theta ))} rd\theta ++\mathbf {\boldsymbol {\hat {\phi ))} r\sin \theta d\phi }$

And,

${\displaystyle du={\frac {\partial u}{\partial r))dr+{\frac {\partial u}{\partial \theta ))d\theta +{\frac {\partial u}{\partial \phi ))d\phi }$

${\displaystyle du=\nabla u\cdot d\mathbf {r} }$

${\displaystyle {\frac {\partial u}{\partial r))dr+{\frac {\partial u}{\partial \theta ))d\theta +{\frac {\partial u}{\partial \phi ))d\phi =\nabla u\cdot d\mathbf {r} }$

Then in spherical coordinates,

${\displaystyle {\frac {\partial u}{\partial r))dr+{\frac {\partial u}{\partial \theta ))d\theta +{\frac {\partial u}{\partial \phi ))d\phi =(\nabla u)_{r}dr+(\nabla u)_{\theta }rd\theta +(\nabla u)_{\phi }r\sin \theta d\phi }$

Therefore,

${\displaystyle (\nabla u)_{r}={\frac {\partial u}{\partial r)),(\nabla u)_{\theta }={\frac {1}{r)){\frac {\partial u}{\partial \theta )),(\nabla u)_{\phi }={\frac {1}{r\sin \theta )){\frac {\partial u}{\partial \phi ))}$

therefore,

${\displaystyle \nabla =\mathbf {\hat {r)) {\frac {\partial }{\partial r))+{\frac {\mathbf {\boldsymbol {\hat {\theta ))} }{r)){\frac {\partial }{\partial \theta ))+{\frac {\mathbf {\boldsymbol {\hat {\phi ))} }{r\sin \theta )){\frac {\partial }{\partial \phi ))}$

${\displaystyle \mathbf {\boldsymbol {\hat {\phi ))} ={\boldsymbol {\hat {\phi ))}=\mathbf {\hat {\phi )) }$

${\displaystyle (A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})}$

${\displaystyle =\left(\mathbf {\hat {r)) {\frac {\partial }{\partial r))+{\frac {\boldsymbol {\hat {\theta ))}{r)){\frac {\partial }{\partial \theta ))+{\frac {\boldsymbol {\hat {\phi ))}{r\sin \theta )){\frac {\partial }{\partial \phi ))\right)\cdot (A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})}$

${\displaystyle =\mathbf {\hat {r)) \cdot {\frac {\partial }{\partial r))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})+{\frac {\boldsymbol {\hat {\theta ))}{r))\cdot {\frac {\partial }{\partial \theta ))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})+{\frac {\boldsymbol {\hat {\phi ))}{r\sin \theta ))\cdot {\frac {\partial }{\partial \phi ))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})}$

${\displaystyle {\frac {\partial }{\partial r))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})=\left({\frac {\partial A_{r)){\partial r))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial r)){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial r)){\boldsymbol {\hat {\phi ))}+A_{r}{\color {red}{\frac {\partial \mathbf {\hat {r)) }{\partial r))}+A_{\theta }{\color {red}{\frac {\partial {\boldsymbol {\hat {\theta )))){\partial r))}+A_{\phi }{\color {red}{\frac {\partial {\boldsymbol {\hat {\phi )))){\partial r))}\right)}$

${\displaystyle =\left({\frac {\partial A_{r)){\partial r))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial r)){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial r)){\boldsymbol {\hat {\phi ))}\right)}$

${\displaystyle {\frac {\partial }{\partial \theta ))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})=\left({\frac {\partial A_{r)){\partial \theta ))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial \theta )){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial \theta )){\boldsymbol {\hat {\phi ))}+A_{r}{\frac {\partial \mathbf {\hat {r)) }{\partial \theta ))+A_{\theta }{\frac {\partial {\boldsymbol {\hat {\theta )))){\partial \theta ))+A_{\phi }{\color {red}{\frac {\partial {\boldsymbol {\hat {\phi )))){\partial \theta ))}\right)}$

${\displaystyle =\left({\frac {\partial A_{r)){\partial \theta ))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial \theta )){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial \theta )){\boldsymbol {\hat {\phi ))}+A_{r}{\boldsymbol {\hat {\theta ))}+A_{\theta }(-\mathbf {\hat {r)) )\right)}$

${\displaystyle {\frac {\partial }{\partial \phi ))(A_{r}\mathbf {\hat {r)) +A_{\theta }{\boldsymbol {\hat {\theta ))}+A_{\phi }{\boldsymbol {\hat {\phi ))})=\left({\frac {\partial A_{r)){\partial \phi ))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial \phi )){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial \phi )){\boldsymbol {\hat {\phi ))}+A_{r}{\frac {\partial \mathbf {\hat {r)) }{\partial \phi ))+A_{\theta }{\frac {\partial {\boldsymbol {\hat {\theta )))){\partial \phi ))+A_{\phi }{\frac {\partial {\boldsymbol {\hat {\phi )))){\partial \phi ))\right)}$

${\displaystyle =\left({\frac {\partial A_{r)){\partial \phi ))\mathbf {\hat {r)) +{\frac {\partial A_{\theta )){\partial \phi )){\boldsymbol {\hat {\theta ))}+{\frac {\partial A_{\phi )){\partial \phi )){\boldsymbol {\hat {\phi ))}+A_{r}\sin \theta {\boldsymbol {\hat {\phi ))}+A_{\theta }\cos \theta {\boldsymbol {\hat {\phi ))}+A_{\phi }[-(\mathbf {\hat {r)) \sin \theta +{\boldsymbol {\hat {\theta ))}\cos \theta )]\right)}$

${\displaystyle {\boldsymbol {\hat {r))}=\mathbf {\hat {r)) }$

aaaaaaaaaaaaaaaa

${\displaystyle =\left({\frac {\partial A_{r)){\partial r))\right)+\left({\frac {1}{r)){\frac {\partial A_{\theta )){\partial \theta ))+{\frac {A_{r)){r))\right)+\left({\frac {1}{r\sin \theta )){\frac {\partial A_{\phi )){\partial \phi ))+{\frac {A_{r)){r))+{\frac {A_{\theta }\cos \theta }{r\sin \theta ))\right)}$

${\displaystyle =\left({\frac {\partial A_{r)){\partial r))+{\frac {2A_{r)){r))\right)+\left({\frac {1}{r)){\frac {\partial A_{\theta )){\partial \theta ))+{\frac {A_{\theta }\cos \theta }{r\sin \theta ))\right)+{\frac {1}{r\sin \theta )){\frac {\partial A_{\phi )){\partial \phi ))}$

${\displaystyle ={\frac {1}{r^{2))}{\frac {\partial }{\partial r))(r^{2}A_{r})+{\frac {1}{r\sin \theta )){\frac {\partial }{\partial \theta ))(A_{\theta }\sin \theta )++{\frac {1}{r\sin \theta )){\frac {\partial A_{\phi )){\partial \phi ))}$

${\displaystyle \mathbf {\hat {r)) \times {\boldsymbol {\hat {\theta ))}={\boldsymbol {\hat {\phi ))},{\boldsymbol {\hat {\theta ))}\times {\boldsymbol {\hat {\phi ))}=\mathbf {\hat {r)) ,{\boldsymbol {\hat {\phi ))}\times \mathbf {\hat {r)) ={\boldsymbol {\hat {\theta ))))$

${\displaystyle \mathbf {\hat {r)) \times {\boldsymbol {\hat {\phi ))}=-{\boldsymbol {\hat {\theta ))},{\boldsymbol {\hat {\theta ))}\times \mathbf {\hat {r)) =-{\boldsymbol {\hat {\phi ))},{\boldsymbol {\hat {\phi ))}\times {\boldsymbol {\hat {\theta ))}=-\mathbf {\hat {r)) }$

cccccccccccccccccccccccccccc

${\displaystyle =\left({\frac {\partial A_{\theta )){\partial r)){\boldsymbol {\hat {\phi ))}-{\frac {\partial A_{\phi )){\partial r)){\boldsymbol {\hat {\theta ))}\right)+}$ ${\displaystyle \left(-{\frac {1}{r)){\frac {\partial A_{r)){\partial \theta )){\boldsymbol {\hat {\phi ))}+{\frac {1}{r)){\frac {\partial A_{\phi )){\partial \theta ))\mathbf {\hat {r)) +{\frac {A_{\theta )){r)){\boldsymbol {\hat {\phi ))}\right)+}$ ${\displaystyle \left({\frac {1}{r\sin \theta )){\frac {\partial A_{r)){\partial \phi )){\boldsymbol {\hat {\theta ))}-{\frac {1}{r\sin \theta )){\frac {\partial A_{\theta )){\partial \phi ))\mathbf {\hat {r)) -{\frac {A_{\phi )){r)){\boldsymbol {\hat {\theta ))}+{\frac {A_{\phi }\cos \theta }{r\sin \theta ))\mathbf {\hat {r)) \right)}$

${\displaystyle ={\frac {\mathbf {\hat {r)) }{r\sin \theta ))\left[{\frac {\partial }{\partial \theta ))(A_{\phi }\sin \theta )-{\frac {\partial A_{\theta )){\partial \phi ))\right]}$${\displaystyle +{\frac {\boldsymbol {\hat {\theta ))}{r\sin \theta ))\left[{\frac {\partial A_{r)){\partial \phi ))-\sin \theta {\frac {\partial }{\partial r))(rA_{\phi })\right]}$${\displaystyle +{\frac {\boldsymbol {\hat {\phi ))}{r))\left[{\frac {\partial }{\partial r))(rA_{\theta })-{\frac {\partial A_{r)){\partial \theta ))\right]}$

lllllllllllllllllllllll

${\displaystyle {\binom {x}{y))=\sum _{k=0}^{n}{\binom {n}{k)){\binom {x-n}{y-k))}$

${\displaystyle {\binom {x}{y))={\binom {1}{0)){\binom {x-1}{y-0))+{\binom {1}{1)){\binom {x-1}{y-1))={\binom {x-1}{y))+{\binom {x-1}{y-1))}$

${\displaystyle {\binom {x}{y))={\binom {x-2}{y))+2{\binom {x-2}{y-1))+{\binom {x-2}{y-2))}$

${\displaystyle {\binom {x}{y))={\binom {x-3}{y))+3{\binom {x-3}{y-1))+3{\binom {x-3}{y-2))+{\binom {x-3}{y-3))}$

${\displaystyle {\binom {x}{y))={\binom {x-4}{y))+4{\binom {x-4}{y-1))+6{\binom {x-4}{y-2))+4{\binom {x-4}{y-3))+{\binom {x-4}{y-4))}$

${\displaystyle \left({\frac {1+{\color {red}2\sin \phi )){2))+{\frac {1}{\color {red}\sin \theta ))={\frac {5}{6))\right)}$

${\displaystyle \nabla ^{2}={\frac {1}{r^{2))}{\color {red}{\frac {\partial }{\partial r))\left(r^{2}{\frac {\partial }{\partial r))\right)}+{\color {blue}{\frac {1}{r^{2}\sin \theta ))}{\frac {\partial }{\partial \theta ))\left(\sin \theta {\frac {\partial }{\partial \theta ))\right)+{\frac {1}{r^{2}\color {green}\sin ^{2}\theta )){\frac {\partial ^{2)){\partial ^{2}\phi ))}$

${\displaystyle (1-x)^{p+1}\sum _{k=0}^{q}{\binom {p+k}{k))x^{k}+x^{q+1}\sum _{k=0}^{p}{\binom {q+k}{k))(1-x)^{k}=1}$

${\displaystyle (1-x)^{n+1}\sum _{k=0}^{m}{\binom {n+k}{k))x^{k}+x^{m+1}\sum _{k=0}^{n}{\binom {m+k}{k))(1-x)^{k}=1}$

${\displaystyle (1-x)^{n+1}\sum _{k=0}^{m}\sum _{i=0}^{k}{\binom {(n-1)+i}{i))x^{k}+x^{m+1}\sum _{k=0}^{n}\sum _{i=0}^{k}{\binom {(m-1)+i}{i))(1-x)^{k}=1}$

${\displaystyle (1-x)^{n+1}\sum _{i=0}^{m}\sum _{k=i}^{m}{\binom {(n-1)+i}{i))x^{k}+x^{m+1}\sum _{i=0}^{n}\sum _{k=i}^{n}{\binom {(m-1)+i}{i))(1-x)^{k}=1}$

${\displaystyle (1-x)^{n+1}\sum _{i=0}^{m}{\binom {(n-1)+i}{i))\sum _{k=i}^{m}x^{k}+x^{m+1}\sum _{i=0}^{n}{\binom {(m-1)+i}{i))\sum _{k=i}^{n}(1-x)^{k}=1}$

${\displaystyle (1-x)^{n+1}\sum _{i=0}^{m}{\binom {(n-1)+i}{i)){\frac {1-x^{m+1)){1-x))+x^{m+1}\sum _{i=0}^{n}{\binom {(m-1)+i}{i)){\frac {1-(1-x)^{n+1)){x))=1}$

${\displaystyle (1-x)^{n}\sum _{i=0}^{m}{\binom {(n-1)+i}{i))(1-x^{m+1})+x^{m}\sum _{i=0}^{n}{\binom {(m-1)+i}{i))(1-(1-x)^{n+1})=1}$

${\displaystyle {\frac {5}{s^{2}(s+1)^{3))}={\frac {A}{s^{2))}+{\frac {B}{s))+{\frac {C}{(s+1)^{3))}+{\frac {D}{(s+1)^{2))}+{\frac {E}{(s+1)))}$

${\displaystyle A=\lim _{s\rightarrow 0}\left[\left({\frac {5}{s^{2}(s+1)^{3))}\right)s^{2}\right]=\lim _{s\rightarrow 0}\left[{\frac {5}{(s+1)^{3))}\right]=5}$

${\displaystyle B=\lim _{s\rightarrow 0}\left[{\frac {d}{ds))\left({\frac {5}{s^{2}(s+1)^{3))}\right)s^{2}\right]=\lim _{s\rightarrow 0}\left[{\frac {d}{ds)){\frac {5}{(s+1)^{3))}\right]==\lim _{s\rightarrow 0}\left[{\frac {5(-3)}{(s+1)^{4))}\right]=-15}$

${\displaystyle C=\lim _{s\rightarrow -1}\left[\left({\frac {5}{s^{2}(s+1)^{3))}\right)(s+1)^{3}\right]=\lim _{s\rightarrow -1}\left[{\frac {5}{s^{2))}\right]=5}$

${\displaystyle D=\lim _{s\rightarrow -1}\left[{\frac {1}{1!)){\frac {d}{ds))\left({\frac {5}{s^{2}(s+1)^{3))}(s+1)^{3}\right)\right]=\lim _{s\rightarrow -1}\left[{\frac {1}{1!)){\frac {d}{ds)){\frac {5}{s^{2))}\right]==\lim _{s\rightarrow -1}\left[{\frac {1}{1!)){\frac {5(-2)}{s^{3))}\right]=10}$

${\displaystyle E=\lim _{s\rightarrow -1}\left[{\frac {1}{2!)){\frac {d^{2)){ds))\left({\frac {5}{s^{2}(s+1)^{3))}(s+1)^{3}\right)\right]=\lim _{s\rightarrow -1}\left[{\frac {1}{2!)){\frac {d^{2)){ds)){\frac {5}{s^{2))}\right]==\lim _{s\rightarrow -1}\left[{\frac {1}{2!)){\frac {5(-2)(-3)}{s^{4))}\right]=15}$

${\displaystyle f(x)={\frac {1}{(1-2x))){\frac {1}{(1-x)^{n+1))))$

${\displaystyle ={\frac {B}{1-2x))+{\frac {A_{0)){(1-x)^{n+1))}+{\frac {A_{1)){(1-x)^{n))}+{\frac {A_{2)){(1-x)^{n-1))}+{\frac {A_{3)){(1-x)^{n-2))}+\cdots +{\frac {A_{n)){(1-x)))}$

${\displaystyle B=\lim _{x\rightarrow 1/2}\left[(1-2x)f(x)\right]={\frac {1}{(1-1/2)^{n+1))}={\frac {1}{(1/2)^{n+1))}=2^{n+1))$

${\displaystyle A_{0}=\lim _{x\rightarrow 1}\left[(x-1)^{n+1}f(x)\right]={\frac {1}{1-2))=-1}$

${\displaystyle A_{1}=\lim _{x\rightarrow 1}\left[{\frac {-1}{1!))\left({\frac {d}{dx))\right)\left({(x-1)^{n+1}f(x)}\right)\right]=\lim _{x\rightarrow 1}\left[{\frac {-1}{1!))\left({\frac {d}{dx))\right){(1-2x)^{-1))\right]=(-1){\frac {(-1)(-2)}{(-1)^{2))}=-2}$

${\displaystyle A_{2}=\lim _{x\rightarrow 1}\left[{\frac {(-1)^{2)){2!))\left({\frac {d}{dx))\right)^{2}\left({(x-1)^{n+1}f(x)}\right)\right]=\lim _{x\rightarrow 1}\left[{\frac {(-1)^{2)){2!))\left({\frac {d}{dx))\right)^{2}{(1-2x)^{-1))\right]={\frac {1}{2)){\frac {8}{(-1)^{3))}=-4.}$

${\displaystyle A_{n}=\lim _{x\rightarrow 1}\left[{\frac {(-1)^{n)){n!))\left({\frac {d}{dx))\right)^{n}{(1-2x)^{-1))\right]=\lim _{x\rightarrow 1}\left[{\frac {(-1)^{n)){n!)){(-1)^{n}n!(1-2x)^{-(n+1)}(-2)^{n))\right]}$

${\displaystyle ={\frac {(-1)^{n)){n!)){(-1)^{n}n!(-1)^{-(n+1)}(-1)^{n}2^{n))=-2^{n}.}$

${\displaystyle {\frac {1}{(1-2x))){\frac {1}{(1-x)^{n+1))}={\frac {2^{n+1)){1-2x))-{\frac {2^{0)){(1-x)^{n+1))}-{\frac {2^{1)){(1-x)^{n))}-{\frac {2^{2)){(1-x)^{n-1))}-{\frac {2^{3)){(1-x)^{n-2))}+\cdots -{\frac {2^{n)){(1-x)))}$

${\displaystyle (1-x)^{p+1}\sum _{k=0}^{q}{\binom {p+k}{p))x^{k}+x^{q+1}\sum _{k=0}^{p}{\binom {q+k}{q))(1-x)^{k}=1}$

${\displaystyle (1-x)^{p+1}+x^{1}\sum _{k=0}^{p}{\binom {k}{0))(1-x)^{k}=1}$

${\displaystyle =\sum _{k=0}^{p}{\binom {q+1+k}{q+1))x(1-x)^{k}+{\binom {q+p+1}{q+1))(1-x)^{p+1}-\sum _{k=0}^{p}{\binom {q+k}{q))(1-x)^{k))$

${\displaystyle =\sum _{k=0}^{p-1}{\binom {q+1+k}{q+1))x(1-x)^{k}+{\binom {q+p+1}{q+1))(1-x)^{p}-\sum _{k=0}^{p}{\binom {q+k}{q))(1-x)^{k))$

${\displaystyle =\sum _{k=0}^{p-1}{\binom {q+1+k}{q+1))x(1-x)^{k}+{\binom {q+p+1}{q+1))(1-x)^{p}-{\binom {q+p}{q))(1-x)^{p}-\sum _{k=0}^{p-1}{\binom {q+k}{q))(1-x)^{k))$

${\displaystyle =\sum _{k=0}^{p-1}{\binom {q+1+k}{q+1))x(1-x)^{k}+{\binom {q+p}{q+1))(1-x)^{p}-\sum _{k=0}^{p-1}{\binom {q+k}{q))(1-x)^{k))$

${\displaystyle =\cdots }$

${\displaystyle ={\binom {q+1}{q+1))x+{\binom {q+1}{q+1))(1-x)-{\binom {q}{q))=0}$

${\displaystyle (1-x)\sum _{k=0}^{q}{\binom {p+k}{p)){\frac {1}{x^{q-k))}+x\sum _{k=0}^{p}{\binom {q+k}{q)){\frac {1}{(1-x)^{p-k))}={\frac {1}{x^{q}(1-x)^{p))))$

${\displaystyle {\frac {1}{\sqrt {1-x))}=1+{\frac {1}{2))x+{\frac {3\cdot 1}{4\cdot 2))x^{2}+{\frac {5\cdot 3\cdot 1}{6\cdot 4\cdot 2))x^{3}+\cdots =\sum {\frac {(2k-1)!!}{(2k)!!))x^{k))$

${\displaystyle {\frac {d}{dx)){\sqrt {1-x))=-{\frac {1}{2)){\frac {1}{\sqrt {1-x))))$

${\displaystyle {\sqrt {1-x))=1-{\frac {1}{2))\left[x+{\frac {1}{2))\left({\frac {1}{2))\right)x^{2}+{\frac {1}{3))\left({\frac {3\cdot 1}{4\cdot 2))\right)x^{3}+{\frac {1}{4))\left({\frac {5\cdot 3\cdot 1}{6\cdot 4\cdot 2))\right)x^{4}+\cdots +{\frac {1}{k+1))\left({\frac {(2k-1)!!}{(2k)!!))\right)x^{k+1}+\cdots \right]}$

${\displaystyle {\sqrt {1-x))={\frac {1-x}{\sqrt {1-x))))$

${\displaystyle {\frac {(2k-1)!!}{(2k)!!))-{\frac {(2k-3)!!}{(2k-2)!!))=-{\frac {1}{2k)){\frac {(2k-3)!!}{(2k-2)!!))}$