User:Gauge00/Equation

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

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

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

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

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

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

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

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

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

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

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

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

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

This 100

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

therefore $$ \sum_{i=0}^{n} \binom{n-i+1}{1} \binom{i}{0} = \binom{n+2}{2} $$

Then

$$ \sum_{i=r}^{n} \binom{n-i+2}{2} \binom{i}{r} = \sum_{i=r}^{n} \sum_{j=0}^{n} \binom{n-i-j+1}{1} \binom{j}{0}  \binom{i}{r} $$

$$= \sum_{i=r}^{n} \sum_{j=0}^{n} \binom{n-i-j+1}{1}\binom{i}{r} $$ $$= \sum_{j=0}^{n} \sum_{i=r}^{n}  \binom{n-i-j+1}{1}\binom{i}{r} = \sum_{j=0}^{n} \binom{n-j+2}{r+2} = \sum_{j=2}^{n+2} \binom{j}{r+2} = \binom{n+3}{r+3}$$

$$ \sum_{i=r}^{n} \binom{n-i+k-1}{k-1} \binom{i}{r} = \binom{n+k}{r+k}$$

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

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

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

Wee see,

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

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

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

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

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

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

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

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

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

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

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

And,

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

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

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

Then in spherical coordinates,

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

Therefore,

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

therefore,

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

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

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

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

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

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

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

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

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

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

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

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

aaaaaaaaaaaaaaaa

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

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

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

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

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

cccccccccccccccccccccccccccc

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

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

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SECTION 100
$$ \binom{x}{y} = \sum_{k=0}^{n} \binom{n}{k} \binom{x-n}{y-k} $$

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

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

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

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

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

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

SECTION 200
$$ (1-x)^{p+1}\sum_{k=0} ^{q} \binom{p+k}{k}x^k + x^{q+1}\sum_{k=0}^{p} \binom{q+k}{k}(1-x)^k =1 $$

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

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

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

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

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

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

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

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

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

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

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

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

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

$$ = \frac{B} {1-2x} + \frac{A_0} {(1-x)^{n+1}} + \frac{A_1} {(1-x)^n} + \frac{A_2} {(1-x)^{n-1}} + \frac{A_3} {(1-x)^{n-2}} + \cdots +\frac{A_n} {(1-x)} $$

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

$$ A_0 = \lim_{x\rightarrow 1} \left[ (x-1)^{n+1}f(x) \right] = \frac{1}{1-2} = -1 $$

$$ A_1 = \lim_{x\rightarrow 1} \left[ \frac{-1}{1!} \left( \frac{d} {dx} \right) \left( {(x-1)^{n+1} f(x)} \right) \right] = \lim_{x\rightarrow 1} \left[ \frac{-1}{1!} \left( \frac{d} {dx} \right) {(1-2x)^{-1}} \right] = (-1) \frac{(-1) (-2)}{(-1)^2} = -2 $$

$$ A_2 = \lim_{x\rightarrow 1} \left[ \frac{(-1)^2}{2!} \left( \frac{d} {dx} \right)^2 \left( {(x-1)^{n+1} f(x)} \right) \right] = \lim_{x\rightarrow 1} \left[ \frac{(-1)^2}{2!} \left( \frac{d} {dx} \right)^2 {(1-2x)^{-1}} \right] = \frac{1}{2} \frac{8}{(-1)^3} = -4. $$

$$ A_n = \lim_{x\rightarrow 1} \left[ \frac{(-1)^n}{n!} \left( \frac{d} {dx} \right)^n {(1-2x)^{-1}} \right] = \lim_{x\rightarrow 1} \left[ \frac{(-1)^n}{n!} {(-1)^n n! (1-2x)^{-(n+1)} (-2)^n} \right] $$

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

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

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

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

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

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

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

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

$$ = \cdots $$

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

ONEONE
$$ (1-x)\sum_{k=0}^q\binom{p+k}{p}\frac{1}{x^{q-k}}+x\sum_{k=0}^p\binom{q+k}{q}\frac{1}{(1-x)^{p-k}}=\frac{1}{x^q(1-x)^p}$$

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

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

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

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

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

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

ONEXX
$$ \rm{area}(sector) = \frac{1}{6}*\rm{area}(disc) = \frac{1}{6} * \frac{\pi}{4} = \frac{\pi}{24} $$

$$ \left(x - \frac{1}{2} \right)^2 + (y - 0)^2 = \left( \frac{1}{2}\right)^2$$

$$ x^2 - x + 1/4 + y^2 = 1$$

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

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

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

$$ y = \sqrt{x(1-x)} = x^{1/2}(1-x)^{1/2} $$

$$ y = x^{1/2}(1-x)^{1/2} $$

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

$$ = x^{1/2} \left(1 - \frac{1}{2}x - \frac{1}{8}x^2 - \frac{1}{16}x^3 - \frac{5}{128}x^4 - \frac{7}{256}x^5 -\cdots \right) $$

$$ = x^{1/2} - \frac{1}{2}x^{3/2} - \frac{1}{8}x^{5/2} - \frac{1}{16}x^{7/2} - \frac{5}{128}x^{9/2} - \frac{7}{256}x^{11/2} -\cdots $$

$$ \int_{x=0}^{1/4} y dx = \int_{x=0}^{1/4} x^{1/2}(1-x)^{1/2} dx $$

$$ = \left( \frac{2}{3} \right) c^{3/2} - \left( \frac{2}{5} \right) \left( \frac{1}{2} \right) c^{5/2} - \left( \frac{2}{7} \right) \left( \frac{1}{8} \right) c^{7/2} - \left( \frac{2}{9} \right) \left( \frac{1}{16} \right) c^{9/2} - \left( \frac{2}{11} \right) \left( \frac{5}{128} \right) c^{11/2} - \left( \frac{2}{13} \right) \left( \frac{7}{256} \right) c^{13/2} -\cdots (c=\frac{1}{4}) $$

$$ = \left( \frac{2}{3} \right) c^{3} - \left( \frac{1}{5} \right) c^{5} - \left( \frac{1}{7\cdot 2^2} \right) c^{7} - \left( \frac{1}{9\cdot 2^3} \right) c^{9} - \left( \frac{5}{11\cdot 2^6} \right) c^{11} - \left( \frac{7}{13\cdot 2^7} \right) c^{13} -\cdots (c=\frac{1}{2}) $$

$$ = \left( \frac{1}{3\cdot 2^2} \right) - \left( \frac{1}{5\cdot 2^5} \right) - \left( \frac{1}{7\cdot 2^9} \right) - \left( \frac{1}{9\cdot 2^{12}} \right) - \left( \frac{5}{11\cdot 2^{17}} \right) - \left( \frac{7}{13\cdot 2^{20}} \right) -\cdots $$

$$ = \frac{1}{12} - \frac{1}{160} - \frac{1}{3584} - \frac{1}{36864} - \frac{5}{1441792} - \frac{7}{13631488} -\cdots $$

$$ = 0.0767732073165$$

$$ \pi = \frac{3}{4} \sqrt{3} + 24 \int y dx $$

$$ = \frac{3}{4} \cdot 1.732050807568877 + 24 \cdot 0.0767732073165 $$

$$ = 3.1415950812734890 $$