Forum Informatyka

Wykład.



(a+b)^{1}=ab(b/a+a/b)

(a+b)^{2}=ab((1+b/a+1+a/b)^{1})(b/a+a/b)/(b/a+a/b)

(a+b)^{3}=ab((1+b/a+1+a/b)(b/a+a/b))^{2}/(b/a+a/b)

(a+b)^{4}=ab((1+b/a+1+a/b)}(b/a+a/b))^{3}/(b/a+a/b)

(a+b)^{5}=ab((1+b/a+1+a/b)(b/a+a/b))^{4}/(b/a+a/b)

(a+b)^{6}=ab((1+b/a+1+a/b)(b/a+a/b))^{5}/(b/a+a/b)

(a+b)^{7}=ab((1+b/a+1+a/b)}(b/a+a/b))^{6}/(b/a+a/b)

Policzmy dla (a+b+c)

(a+b+c)^{1}=(ab+ac+bc)(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{2}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)^{1})+(b/a+a/b+a/c+c/a+b/c+c/b)/(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{3}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)(b/a+a/b+a/c+c/a+b/c+c/b))^{2}/(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{4}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)}(b/a+a/b+a/c+c/a+b/c+c/b))^{3}/(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{5}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)(b/a+a/b+a/c+c/a+b/c+c/b))^{4}/(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{6}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)(b/a+a/b+a/c+c/a+b/c+c/b))^{5}/(b/a+a/b+a/c+c/a+b/c+c/b)

(a+b+c)^{7}=(ab+ac+bc)((1+b/a+1+a/b+1+a/c+c/a+b/c+c/b)}(b/a+a/b+a/c+c/a+b/c+c/b))^{6}/(b/a+a/b+a/c+c/a+b/c+c/b)

Liczmy dalej:



(b/a+a/b))^{2}=1+a^{2}/b^{2}+1+b^{2}/a^{2}

(b/a+a/b))^{3}=a/b+a/b+b/a+b/a

(b/a+a/b))^{4}=2*(1+a^{2}/b^{2}+1+b^{2}/a^{2})

(b/a+a/b))^{5}=2*(2a/b+2b/a)

(b/a+a/b))^{6}=4*(1+a^{2}/b^{2}+1+b^{2}/a^{2})

(b/a+a/b))^{7}=4*(2a/b+2b/a)

Zastanawiam się czy liczyć działania.

Przy okazji wychodzi waga, prądowa, policzę tylko, jeden przykład, reszta jutro. Bo jestem już po lekach.

(a+b)^{3}=ab((1+b/a+1+a/b)(b/a+a/b))=ab((1+b/a+1+a/b)(1+a^{2}/b^{2}+1+b^{2}/a^{2})/(b/a+a/b)