Question

1+2¹+2²+2³+2⁴+...+2ⁿ•2=131 071

Answer 1

daca n este nr natural si 2 este inmultit numai cu 2 la puterea n se rezolva astfel
1+2+2²+................... + $2^{n+1}$=131071
avem suma primilor n+1 termeni ai unei progresii geometrice
$s_{n+1}$=$b_{1}$·(1-$q^{n+1}$)/(1-q) =131071
1·(1-$2^{n+1}$)/(1-2)=131071
$2^{n+1}$ = 131072=$2^{17}$
n+1=17
n=16