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$$\left(a\pm b\right)^2=a^2\pm 2ab+b^2\quad\left(a\pm b\right)^3=a^3\pm 3a^2b+3ab^2\pm b^3\quad\left(a+b\right)^n=\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}\quad a^2-b^2=\left(a-b\right)\left(a+b\right)\quad a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$$
$$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)\quad a^4+4b^4=\left(a^2+2ab+2b^2\right)\left(a^2-2ab+2b^2\right)\quad\left(a^2+nb^2\right)\left(c^2+nd^2\right)=\left(ac\pm nbd\right)^2+n\left(ad\mp bc\right)^2$$
$$a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\quad ax^2+bx+c=0\iff x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\quad D_2=b^2-4ac\quad\alpha_2+\beta_2=-\frac{a}{b}\quad\alpha_2\beta_2=\frac{c}{a}$$
$$y^3+ay+b=0\iff y=\frac{1}{3}\left(\sqrt[3]{-\cfrac{27}{2}b+\cfrac{3\sqrt{3}i}{2}\sqrt{-4a^3-27b^2}}-\cfrac{3a}{\sqrt[3]{-\cfrac{27}{2}b+\cfrac{3\sqrt{3}i}{2}\sqrt{-4a^3-27b^2}-}}\right)\quad(y\to x-\frac{b}{3})$$
$$D_3=-4ac^3+b^2c^2-4b^2d+18abcd-27a^2d^2\quad\alpha_3+\beta_3+\gamma_3=-\frac{b}{a}\quad\alpha_3\beta_3+\beta_3\gamma_3+\gamma_3\alpha_3=\frac{c}{a}\quad\alpha_3\beta_3\gamma_3=-\frac{d}{a}\quad\frac{a+b}{2}\geq\sqrt{ab}\quad a^2+b^2+c^2\geq ab+bc+ca$$
$$\sum_{i=1}^na_i\geq n\sqrt[n]{\prod_{i=1}^na_i}\quad\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}\quad a\sqrt{x}+b\sqrt{y}\leq\sqrt{\left(a^2+b^2\right)\left(x+y\right)}\quad\log x\leq x-1\quad\left(1+x\right)^n\geq1+nx\quad\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)\geq\left(a_1b_1+a_2b_2\right)^2$$
$$\left(a_1^2+a_2^2+a_3^2\right)\left(b_1^2+b_2^2+b_3^2\right)\geq\left(a_1b_1+a_2b_2+a_3b_3\right)^2\quad\left(\sum_{i=1}^na_i^2\right)\left(\sum_{i=1}^nb_i^2\right)\geq\left(\sum_{i=1}^na_ib_i\right)^2\quad\frac{\pi}{2}x\leq\sin x\leq x(0\geq x\geq\frac{\pi}{2})\quad\sin^2\theta+\cos^2\theta=1\quad 1+\tan^2\theta=\frac{1}{\cos^2\theta}$$
$$\sin\left(a\pm b\right)=\sin a\cos b\pm\cos a\sin b\quad\cos\left(a\pm b\right)=\cos a\cos b\mp\sin a\sin b\quad\tan\left(a\pm b\right)=\frac{\tan a\pm\tan b}{1\mp\tan a\tan b}\quad\sin2\theta=2\sin\theta\cos\theta\quad\cos2\theta=2\cos^2\theta-1$$
$$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}\quad\sin3\theta=-4\sin^3\theta+3\sin\theta\quad\cos3\theta=4\cos^3\theta-3\cos\theta\quad\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\quad\sin^2\theta=\frac{1-\cos2\theta}{2}\quad\cos^2\theta=\frac{1+\cos2\theta}{2}$$
$$\quad\tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta}\quad e^{i\theta}=\cos\theta+i\sin\theta\quad\cosh^2\theta-\sinh^2\theta=1\quad 1-\tanh^2\theta=\frac{1}{\cosh^2\theta}\quad\sinh\left(a\pm b\right)=\sinh a\cosh b\pm\cosh a\sinh b$$
$$\cosh\left(a\pm b\right)=\cosh a\cosh b\pm\sinh a\sinh b\quad\tanh\left(a\pm b\right)=\frac{\tanh a\pm\tanh b}{1\pm\tanh a\tanh b}\quad \sinh2x=2\sinh x\cosh x\quad\cosh2x=2\cosh x^2-1\quad\tanh2x=\frac{2\tanh x}{1+\tanh^2x}$$
$$\sinh3x=4\sinh x^3+3\sinh x\quad\cosh3x=4\cosh^3x-3\cosh x\quad\tanh3x=\frac{3\tanh x+\tanh^3x}{1+3\tanh^2x}\quad\sinh x=-i\sin(ix)\quad\cosh x=\cos(ix)$$
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\quad a=b\cos C+c\cos B\quad a^2=b^2+c^2-2bc\cos A\quad\cos A=\frac{b^2+c^2-a^2}{2bc}\quad\frac{a-b}{a+b}=\frac{\tan\left(\frac{1}{2}\left(A-B\right)\right)}{\tan\left(\frac{1}{2}\left(A+B\right)\right)}\quad\cot\left(\frac{A}{2}\right)=\frac{s-a}{\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}}$$
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{1}{4}\left(\cos A+\cos B+\cos C-1\right)=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{abc}\quad\sin^2A+\sin^2B+\sin^2C=2\left(\cos A\cos B\cos C+1\right)$$
$$\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\frac{1}{4}\left(\sin A+\sin B+\sin C\right)=\frac{s}{4R}\quad\tan A\tan B\tan C=\tan A+\tan B+\tan C\quad\frac{1}{\tan A_{/2}\tan B_{/2}\tan C_{/2}}=\frac{1}{\tan A_{/2}}+\frac{1}{\tan B_{/2}}+\frac{1}{\tan C_{/2}}$$
$$r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\quad S=\frac{1}{2}bc\sin A=rs=\frac{abc}{4R}=2R^2\sin A\sin B\sin C=\frac{a^2\sin B\sin C}{2\sin A}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}=r_A\left(s-a\right)=\sqrt{rr_Ar_Br_C}$$
$$\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x\to a}\frac{f'\left(x\right)}{g'\left(x\right)}\quad f'\left(a\right)=\lim_{h\to 0}\frac{f\left(a+h\right)-f\left(a\right)}{h}\quad\left(fg\right)'=f'g+fg'\quad\left(fg\right)''=f''g+2f'g'+fg''\quad\left(fg\right)^{\left(n\right)}=\sum_{k=0}^{n}\binom{n}{k}f^{\left(k\right)}g^{\left(n-k\right)}\quad\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}$$
$$\left(\frac{g}{f}\right)'=\frac{g'f-f'g}{f^2}\quad\left(f\circ g\right)'=f'\circ gg'\quad \left(x^a\right)'=ax^{a-1}\quad\left(\sin x\right)'=\cos x\quad\left(\cos x\right)'=-\sin x\quad\left(\tan x\right)'=\frac{1}{\cos^2x}\quad\left(\cot x\right)'=-\frac{1}{\sin^2x}\quad\left(a^x\right)'=a^x\log a$$
$$\left(\log_ax\right)'=\frac{1}{x\log a}\quad\left(\arcsin x\right)'=\frac{1}{\sqrt{1-x^2}}\quad\left(\arccos x\right)'=-\frac{1}{\sqrt{1-x^2}}\quad\left(\arctan x\right)'=\frac{1}{1+x^2}\quad\left(\sinh x\right)'=\cosh x\quad\left(\cosh x\right)'=\sinh x\quad\left(\tanh x\right)'=1-\tanh^2x$$
$$\left(\log x\right)^{(n)}=-\left(n-1\right)!\left(-x\right)^{-n}\quad\int\left(x-a\right)^t~dx=\frac{1}{t+1}\left(x-a\right)^{t+1}+C\quad\int_\alpha^\beta\left(x-\alpha\right)\left(\beta-x\right)~dx=\frac{\left(\beta-\alpha\right)^3}{6}\quad\int_a^bf\left(x\right)~dx=\frac{b-a}{6}\left(f\left(a\right)+4f(\frac{a+b}{2})+f\left(b\right)\right)$$
$$\int_0^1x^m\left(1-x\right)^n~dx=\frac{m!n!}{\left(m+n+1\right)!}\quad\int\frac{1}{\sin x}~dx=\frac{1}{2}\log\left(\frac{1-\cos x}{1+\cos x}\right)+C\quad\int\frac{1}{\cos x}~dx=\frac{1}{2}\log\left(\frac{1+\sin x}{1-\sin x}\right)+C\quad\int\frac{1}{\tan x}~dx=\log\left|\sin x\right|+C$$
$$\int e^{ax}\sin bx~dx=\frac{e^{ax}}{a^2+b^2}\left(a\sin bx-b\cos bx\right)+C\quad\int e^{ax}\cos bx~dx=\frac{e^{ax}}{a^2+b^2}\left(a\cos bx+b\sin bx\right)+C\quad\int\frac{1}{\sqrt{x^2+a^2}}~dx=\log\left(x+\sqrt{x^2+a^2}\right)+C$$
$$\int\sqrt{x^2+a^2}~dx=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\log\left(x+\sqrt{x^2+a^2}\right)\right)+C\quad\int \frac{1}{x^2-a^2}~dx=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+C\quad\int_{-\infty}^\infty e^{-ax^2}~dx=\sqrt{\frac{\pi}{a}}\quad \int_0^\infty\frac{x^3}{e^x-1}~dx=\frac{\pi^4}{15}$$
$$\int_0^\infty x^{2n+1}e^{-ax^2}~dx=\frac{n!}{2a^{n+1}}\quad\int_{-\infty}^\infty x^{2n}e^{-ax^2}~dx=\frac{\left(2n-1\right)!!}{2^na^n}\sqrt{\frac{\pi}{a}}\quad\int_{-\infty}^\infty e^{-ax^2+bx+c}~dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}+c}\quad\int_{-\infty}^\infty \sin^2x~dx=\int_{-\infty}^\infty \cos^2x~dx=\sqrt{\frac{\pi}{2}}$$
$$S=\int_a^b\sqrt{1+f'\left(x\right)^2}~dx=\int_\alpha^\beta\frac{1}{2}\left(xy'-yx'\right)~dt=\int_\alpha^\beta\frac{1}{2}r\left(\theta\right)^2d\theta\quad L=\int_\alpha^\beta\sqrt{f'\left(t\right)^2+g'\left(t\right)^2}~dt=\int_\alpha^\beta\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}~d\theta$$
$$V=\int_a^b\pi\left(f\left(x\right)\right)^2~dx=\frac{2}{3}\pi\int_\alpha^\beta r\left(\theta\right)^3\sin\theta ~d\theta=\int_a^b2\pi x\left|f\left(x\right)\right|~dx=\pi\cos\theta\int_a^b\left(mx-f\left(x\right)\right)^2~dx\quad\Gamma\left(x\right)=\int_0^\infty t^{x-1}e{-t}~dt$$
$$\zeta\left(s\right)=\sum_{n=1}^\infty\frac{1}{n^s}\quad\sinh\left(x\right)=\frac{e^x-e^{-x}}{2}\quad\cosh\left(x\right)=\frac{e^x+e^{-x}}{2}\quad\tanh\left(x\right)=\frac{e^x-e^{-x}}{e^x+e^{-x}}\quad \mathrm{gd} \left(x\right)=\int_{0}^{x}\frac{1}{\cosh t}~dt\quad \varsigma_a\left(x\right)=\frac{1}{1+e^{-ax}}=\frac{\tanh\left(\frac{ax}{2}\right)+1}{2}$$
$$F\left(x,k\right)=\int_0^x\frac{1}{\sqrt{\left(1-t^2\right)\left(1-k^2 t^2\right)}}~dt\quad E\left(x,k\right)=\int_0^x\sqrt{\frac{1-k^2 t^2}{1-t^2}}~dt\quad\Pi\left(a;x,k\right)=\int_0^x\frac{1}{\left(1-at^2\right)\sqrt{\left(1-t^2\right)\left(1-k^2 t^2\right)}}~dt\quad J_n\left(x\right)=\frac{1}{\pi i^n}\int_0^\pi e^{ix\cos\theta}\cos n\theta~d\theta$$
$$\sum_{k=1}^nk=\frac{1}{2}n\left(n+1\right)\quad\sum_{k=1}^nk^2=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)\quad\sum_{k=1}^nk^3=\left(\frac{1}{2}n\left(n+1\right)\right)^2\quad\sum_{k=1}^nk^a=\frac{1}{a+1}\sum_{k=0}^a\binom{a+1}{k}B_kn^{a-k+1}\quad\prod_{k=1}^na^{a_k}=a^{\sum_{k=1}^na_k}$$
$$\prod_{k=1}^n\left(k+a\right)=\frac{\left(a+n\right)!}{a!}\quad F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)\quad V_n=\frac{\pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}+1\right)}\quad e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\quad\sin x=\sum_{k=0}^\infty\left(-1\right)^k\frac{x^{2k+1}}{(2k+1)!}\quad\cos x=\sum_{k=0}^\infty\left(-1\right)^k\frac{x^{2k}}{(2k)!}$$
$$\tan x=\sum_{k=1}^\infty\frac{B_{2k}\left(-4\right)^k\left(1-4^k\right)}{\left(2k\right)!}x^{2k-1}\quad\log\left(x+1\right)=\sum_{k=0}^\infty\left(-1\right)^k\frac{x^{k+1}}{k+1}\quad\frac{1}{\left(1-ax\right)^m}=\sum_{k=0}^\infty\binom{m+k-1}{m-1}a^kx^k\quad W_0\left(x\right)=\sum_{k=0}^\infty\frac{\left(-k\right)^{k-1}}{k!}x^k$$
$$\sinh x=\sum_{k=0}^\infty\frac{1}{\left(2k+1\right)!}x^{2k+1}\quad\cosh x=\sum_{k=0}^\infty\frac{1}{\left(2k\right)!}x^{2k}\quad\tanh x=\sum_{k=1}^\infty\frac{B_{2k}4^k\left(4^k-1\right)}{\left(2k\right)!}x^{2k-1}\quad\sum_{k=1}^\infty k=-\frac{1}{12}\quad\sum_{k=1}^\infty F_k=-1\quad\sum_{k=0}^n\binom{p}{k}\binom{q}{n-k}=\binom{p+q}{n}$$
$$Ri_n=\binom{n}{4}+\frac{-5n^3+45n^2-70n+24}{24}\delta_2\left(n\right)-\frac{3n}{2}\delta_4\left(n\right)+\frac{-45n^2+262n}{6}\delta_6\left(n\right)+42n\delta_{12}\left(n\right)+60n\delta_{18}\left(n\right)+35n\delta_{24}\left(n\right)-38n\delta_{30}\left(n\right)-82n\delta_{42}\left(n\right)-330n\delta_{60}\left(n\right)-144n\delta_{84}\left(n\right)-96n\delta_{90}\left(n\right)-144n\delta_{120}\left(n\right)-96n\delta_{210}\left(n\right)$$
$$\sum_{n=1}^\infty\frac{1}{n^s}=\prod_{p:prime}\cfrac{1}{1-\cfrac{1}{p^s}}\quad\sum_{n=1}^\infty\frac{\lambda\left(n\right)}{n^s}=\frac{\zeta\left(2s\right)}{\zeta\left(s\right)}\quad\sum_{n=1}^\infty\frac{\mu\left(n\right)}{n^s}=\frac{1}{\zeta\left(s\right)}\quad\pi=4\sum_{n=0}^\infty\frac{\left(-1\right)^n}{2n+1}=2\sqrt{3}\sum_{n=0}^\infty\frac{\left(-1\right)^n}{3^n\left(2n+1\right)}=\prod_{n=1}^\infty\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right)$$
$$\frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{n=0}^\infty\frac{\left(26390n+1103\right)\cdot\left(4n\right)!}{\left(4^n99^n\cdot n!\right)^4}=\frac{12}{640320\sqrt{640320}}\sum_{n=0}^\infty\frac{\left(545140134n+13591409\right)\cdot\left(6n\right)!}{\left(-640320\right)^{3n}\cdot\left(3n\right)!\cdot\left(n!\right)^3}\quad\frac{4}{\pi}=\sum_{n=0}^\infty\frac{\left(-1\right)^n\left(21460n+1123\right)\cdot\left(4n\right)!}{882^{2n+1}\left(4^nn!\right)^4}$$
$$\frac{1}{\pi}=\frac{12}{\left(1249638720+159999840\sqrt{61}\right)\sqrt{1249638720+159999840\sqrt{61}}}\sum_{n=0}^{\infty} \frac{(-1)^{n}(6n)!(1657145277365+212175710912\sqrt{61}+\left(107578229802750+3773980892672\sqrt{61}\right)n)}{(3n)!(n!)^3 \left(1249638720+159999840\sqrt{61}\right)^n}$$