Question

calculati:3x(1+2^1+2^2+......+2^49):(1+4^1+4^2+....+4^24)

Answer 1

Avem sume de doua progresi geometrice,Prima cu ratia 2 si 50 termeni.  Formula sumei este $S_{n}= b_{1} \frac{ q^{n}-1 }{q-1},deci:1+2^1+2^2+...+ 2^{49}=1* \frac{ 2^{50}-1 }{2-1}= 2^{50}-1$
A doua suma, cu ratia 4si 25 termeni n=25:
$1+4^1+4^2+...+ 4^{24}= \frac{ 4^{25}-1 }{4-1}= \frac{1}{3}*( 4^{25}-1)= \frac{1}{3}( 2^{50}-1)$ 
Deci avem de facut impartirea: $3*( 2^{50}-1): \frac{ 2^{50}-1 }{3} =3 ( 2^{50}-1)* \frac{3}{( 2^{50}-1)}=9$