Question

Sub l,1 si 2,exercitiile cu combinari,aranjamente si suma,va rog este urgent

Answer 1

[tex]1. C ^{4} _{5} +A ^{4} _{5} C ^{4} _{5}= \frac{5!}{4!*1!} =5 A ^{4} _{5} =\frac{5!}{1!} =120  
5+120=125 [/tex] Am folosit formula Combinarilor si Aranjamentelor, iar apoi am simplificat.

$1+ \frac{1}{3} + \frac{1}{3 ^{2} } + \frac{1}{ 3^{3} } + \frac{1}{ 3^4} } =1+ \frac{3 ^{4}+3^3+3^2+3+1 }{3^4} = \frac{81+27+9+3+1}{81} = \frac{121}{81}$. Am adus la acelasi numitor, acesta fiind 3^4, iar apoi am ridicat la putere.

Answer 2

$\displaystyle 1).C_5^4+A_5^4 \\ \\ \boxed{C_n^k= \frac{n!}{k!(n-k)! } }~~~~~~~~~~~~~~~~~~~~~\boxed{A_n^k= \frac{n!}{(n-k)!} }\\ \\ C_5^4= \frac{5!}{4!(5-4)! } = \frac{5!}{4! \cdot 1!} = \frac{\not4! \cdot 5}{\not 4! \cdot 1} = \frac{5}{1}=5 \\ \\ A_5^4= \frac{5!}{(5-4)!} = \frac{5!}{1!} = \frac{\not1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}{\not1} =2 \cdot 3 \cdot 4 \cdot 5=120 \\ \\ C_5^4+A_5^4=5+120=125$
$\displaystyle 2).1+ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} =1+ \frac{1}{3} + \frac{1}{9} +\frac{1}{27} + \frac{1}{81} = \\ \\ = \frac{81+27+9+3+1}{81} = \frac{121}{81}$