Arithmetics

$$\left({a±b}\right)^2=a^2±2ab+b^2\qquad \left({a±b}\right)^3=a^3±3a^2b+3ab±b^3$$

$$a^2-b^2=\left({a-b}\right)\left({a+b}\right)\qquad a^3±b^3=\left({a±b}\right)\left({a^2∓ab+b^2}\right)$$

$$...$$

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Series/Progression :

$$Def.\ :\ A_n=A_1+\left({n-1}\right)·d\\ S_n=\frac{n·\left({A_1+A_n}\right)}2=\frac{\left({2·A_1+\left({n-1}\right)·d}\right)·n}2\\ \displaystyle{S_{n_3}=\left({A_{n_2}+A_{n_1}+\left({A_{n_2}-A_{n_1}}\right)\frac{n_3-n_2-n_1+1}{n_2-n_1}}\right)·\frac{n_3}2}$$

 

$$Def\ :\ A_n=A_1·q^{n-1}\\ S_n=A_1·\frac{q^n-1}{q-1}=A_1·\frac{1-q^n}{1-q}\\ \lim_{n\ →\ 1,0,-1...-∞}S=\frac{A_1}{1-\frac1q}\ ,\ \frac{\frac12}{1-\frac12}→1$$

 

$$\displaystyle{\boxed{\lim_{n\ →\ ∞}\left({1±\frac xn}\right)^n=e^{±x}=\sum_{i=0}^{n\ →\ ∞}\frac{x^i}{i\ !}}}$$

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$$\displaystyle{\sum_{i\ =\ 1}^n\ \left({2i-1}\right)\ =\ n^2\ =\ 1+3+5+7+9+11+...\\ \sum_{i\ =\ 1}^n\ 2i\ =\ n· \left({n+1}\right)\ =\ \left({0+}\right)2+4+6+8+10+12+...\\ odd\ :\ \sum_{i\ =\ 1}^m\ i\ =\ \sum_{j\ =\ 1}^{n=\frac{m+1}2}\ \left({2j-1}\right)\ +\ \sum_{k\ =\ 1}^{N=\frac{m-1}2}\ k\ =\ \left({\frac{m+1}2}\right)^2+\left({\frac{m-1}2· \left({\frac{m-1}2+1}\right)}\right)\ =\\ =\ \frac{m^2\cancel{+2m}\bcancel{+1}}4 + \frac{m^2\cancel{-2m}\bcancel{+1}+2m\bcancel{-2}}4=\frac{2m^2+2m}4=\frac{m^2+m}2\\ even\ :\ \sum_{i\ =\ 1}^m\ i\ =\ \sum_{j\ =\ 1}^{n=\frac m2}\ \left({2j-1}\right)\ +\ \sum_{k\ =\ 1}^{N=\frac m2}\ k\ =\ \left({\frac m2}\right)^2+\left({\frac m2\left({\frac m2+1}\right)}\right)=\\ =\frac{2m^2+2m}4=\frac{m^2+m}2\\ \sum_{i\ =\ 1}^n\ i\ =\ \frac{n^2+n}2\ =\ \frac{n· \left({n+1}\right)}2}$$

$$\displaystyle{\sum_{i\ =\ 1}^n\ i^2\ =\ \frac{2n^3+3n^2+n}6\ =\ \frac{n^3}3+\frac{n^2}2+\frac n6\ =\ \frac{n\left({n\left({2n+3}\right)+1}\right)}6\ =\ \frac{n\left({n+1}\right)\left({2n+1}\right)}6}$$

$$\displaystyle{\sum_{i\ =\ 1}^n\ i^3\ =\ \left[{\frac{n\left({n+1}\right)}2}\right]^2}$$

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dd

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[Eop]