Question

Se considera functia f:R->R, f(x)=x+1. Calculati suma f(2)+f(2^2)+...+f(2^9);

Answer 1

$f:\mathbb{R} \to \mathbb{R}~~,f(x) =x+1 \\ \\ \\ f(2^1) = 2+1 \\ \\ f(2^2) = 4+1 \\ . \\ . \\ . \\ \\ f(2^9) = 512 + 1 \\ \\ \\ 1+1+ .... +1 = 9 ~\to \text {Deoarece sunt 9 termeni cu 1.} \\ \\ 2+4+ ...+512 = 2(1+2+ ... +256) = 2 \cdot \dfrac {256(256+1)}{2} = 256 \cdot 257 \\ \\ \\ Deci \Rightarrow \boxed {\boxed {S = 256 \cdot 257 + 9 }}$