Question

Sejam $P=\sum\limits_{p=0}^{50} \binom {101} {2p}$ e $Q=\sum\limits_{q=0}^{100} \binom {201} {2q+1}$. Concluímos que o valor de $\frac{log_4 \sqrt[25]{P*Q} }{3}$ é:
a)1
b)2
c)3
d)4
e)5

Answer 1

Ola Joaovictor

P = (0,50)∑ C(101,2p)
P = 1267650600228229401496703205376

Q = (0,100)∑ C(201,2q+1) Q = 1606938044258990275541962092341162602522202993782792835301376

log4(P*Q) = 150 

log4(²⁵√P*Q)/3 = 150/75 = 2 (B)