Question

Fie f:R cu valori in R, f(x)=x+2. Sa se calculeze f(2)+f(2^2)+...+f(2^7).

Answer 1

[tex]S=(2+2)+(2^{2}+2)+(2^{3}+2)+...+(2^7+2)\\ S=2^{1}+2^{2}+...+2^{7}+2*7\\ S_{1}=2^{1}+2^{2}+...+2^{7}\\ 2S_{1}=2^{2}+2^{3}+...2^{7}+2^{8}\\ 2S_{1} =S_{1}-2+2^{8}==\ \textgreater \ S_{1}=2^{8}-2\\ S =S_{1}+2*7 =2^{8}+14-2=2^{8}+12=268[/tex]

Answer 2

f(2)=2+2=4
f(4)=4+2=6
f(8)=8+2=10
f(16)=16+2=18
f(32)=32+2=34
f(64)=64+2=68
f(128)=128+2=130

=4+6+10+18+34+68+68+130
=338