Question

Se considera functia f:R->R, f(x)=2+x. Sa se calculeze f(1)+f(2)+...+f(20)

Answer 1

f(1) = 3
f(2) = 4
.
.
.
f(20) = 22

1 + 2+ 3 +4+.....+ 22
                 ↓
S= n(n+1)/2 = 22×23/2 = 253

⇒ f(1) +f(2) +....f(20) = 253 -3 = 250

Answer 2

f(1 ) + f(2 ) + ... + f( 20 ) = 2 · 20 +  ( 1+2+3 .. +20 ) =
= 40  + ( 1 +20 ) ·20 / 2 = 40 + 21·10 = 40 +210  = 250