Question

7)Sa se rezolve ecuatiile:
a)$2 \frac{(n+5)!}{n!} = \frac{(n+6)!}{(n+1)!}$
b)$\frac{ P_{n} }{ P_{n-1} } = \frac{ P_{12} }{ P_{10} }$

Answer 1

a) 2(n+5)!/n! = (n+6)!/(n+1)!
2n!(n+1)(n+2)(n+3)(n+4)(n+5)/n! = (n+1)!(n+2)(n+3)(n+4)(n+5)(n+6)/(n+1)!
2(n+1)(n+2)(n+3)(n+4)(n+5) = (n+2)(n+3)(n+4)(n+5)(n+6)
2(n+1) = n+6 => 2n+2=n+6 => n=4

b) P(n)/P(n-1) = P(12)/P(10)
P(n) = n!

n!/(n-1)! = 12!/10!
(n-1)!n/(n-1)! = 10!*11*12/10!
n=11*12=132