Question

f(x)=2^x+x. Să se calculeze f(1)+f(2)+⋯+f(8)

Answer 1

Explicație pas cu pas:

f(1)=2^1+1=3

f(2)=2^2+2=6

f(8)=2^8+8=264

f(1)+f(2)+...+f(8)=(2^1+1)+(2^2+2)+...+(2^8+8)=

=(2^1+2^2+...+2^8)+(1+2+...+8)=2^(8+1)-2+(8*9)/2=2^9-2+36=546

Answer 2

Răspuns:

546

Explicație pas cu pas:

$f(1)+f(2)+...+f(8)=(2^1+1)+(2^2+2)+(2^3+3)+...+(2^8+2)=\\ \\=(2^1+2^2+2^3+...+2^8)+(1+2+3+...+8)=\\ \\=(2^9-2)+\frac{8\cdot 9}{2}=\\ \\=512-2+4\cdot 9=\\ \\=510+36=\\ \\=546$