Question

31x2 + 62x$2^{5}$ +62 x$2^{10}$ + 62 x $2^{15}$ + ... + 31 x $2^{n+1}$ + $31^{0}$ x 2 = $16^{50}$ x 2

Answer 1

62(1+2^5+2^10+....+2^n) +2=16^50*2=(2^4)^50*2=2^201
S=1+2^5+2^10+.....2^n
Este o progresie geometrica cu a1=1, ratia q=2^5 si an=2^n
Suma este $\frac{a1-an*q}{1-q}$=$\frac{1-2^n*2^5}{1-32}$= - $\frac{1-2^(n+5)}{31}$
62 *  [- $\frac{1-2^(n+5)}{31}$] +2 =2^201

2^(n+6)-2+2=2^201
2^(n+6)  = 2^201 ⇒ n+6=201⇒ n=195