Resolver S=a(1-r^n)/1-r

$$S = \frac { a ( 1 - r ^ { n } ) } { 1 - r }$$

Ebatzi: a

$$\left\{\begin{matrix}a=-\frac{S\left(r-1\right)}{1-r^{n}}\text{, }&\left(r=0\text{ and }n>0\right)\text{ or }\left(n\neq 0\text{ and }r\neq -1\text{ and }Denominator(n)\text{bmod}2=1\text{ and }r<0\right)\text{ or }\left(r<0\text{ and }Numerator(n)\text{bmod}2=1\text{ and }Denominator(n)\text{bmod}2=1\right)\text{ or }\left(n\neq 0\text{ and }r\neq 1\text{ and }r>0\right)\\a\in \mathrm{R}\text{, }&r\neq 1\text{ and }r\neq 0\text{ and }\left(n=0\text{ or }r<0\right)\text{ and }\left(r=-1\text{ or }n=0\right)\text{ and }Numerator(n)\text{bmod}2=0\text{ and }Denominator(n)\text{bmod}2=1\text{ and }S=0\end{matrix}\right.$$

Ekuazio lineala ebazteko urratsak

Biderkatu ekuazioaren bi aldeak honekin: -r+1.

$$S\left(-r+1\right)=a\left(1-r^{n}\right)$$

Erabili banaketa-propietatea S eta -r+1 biderkatzeko.

$$-Sr+S=a\left(1-r^{n}\right)$$

Erabili banaketa-propietatea a eta 1-r^{n} biderkatzeko.

$$-Sr+S=a-ar^{n}$$

Trukatu aldeak, aldagaiak ezkerraldean egon daitezen.

$$a-ar^{n}=-Sr+S$$

Konbinatu a duten gai guztiak.

$$\left(1-r^{n}\right)a=-Sr+S$$

Modu arruntean dago ekuazioa.

$$\left(1-r^{n}\right)a=S-Sr$$

Zatitu ekuazioaren bi aldeak 1-r^{n} balioarekin.

$$\frac{\left(1-r^{n}\right)a}{1-r^{n}}=\frac{S-Sr}{1-r^{n}}$$

1-r^{n} balioarekin zatituz gero, 1-r^{n} balioarekiko biderketa desegiten da.

$$a=\frac{S-Sr}{1-r^{n}}$$

Zatitu -Sr+S balioa 1-r^{n} balioarekin.

$$a=\frac{S\left(1-r\right)}{1-r^{n}}$$

Ebatzi: S

$$S=-\frac{a\left(r^{n}-1\right)}{1-r}$$

$$\left(r>0\text{ and }r\neq 1\right)\text{ or }\left(n>0\text{ and }r=0\right)\text{ or }\left(Denominator(n)\text{bmod}2=1\text{ and }r<0\right)$$