Fie functia f : R -> R, f(x) = 2x-1. Calculati suma S= f(1) + f(2) +f(3) +.....+f(2012)
Stiu ca numerele cresc din 2 in 2.
f(1)=1
f(2)=3
f(3)=5
f(4)=7 si tot asa.
As vrea sa stiu cum procedez la astfel de exercitii in general . Multumesc :) .

f(1)=2x1-1
f(2)=2x2-1
...............
f(2012)=2x2012-1
________________
S=2(1+2+3+...+2012)-1x2012=[tex]=2\cdot \frac{(1+2012)\cdot2012}{2} -2012=2013\cdot 2012-2012=2012(2013-1)=\\ =2012^2=4048144[/tex]

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Juliajuliana1id Juliajuliana1id · Answer 2

Juliajuliana1id Juliajuliana1id

f(1)=2x1-1
f(2)=2x2-1
...............
f(2012)=2x2012-1
________________
S=2(1+2+3+...+2012)-1x2012=

Answer Link

Fie functia f : R -> R, f(x) = 2x-1. Calculati suma S= f(1) + f(2) +f(3) +.....+f(2012)Stiu ca numerele cresc din 2 in 2.f(1)=1f(2)=3f(3)=5f(4)=7 si tot asa.As vrea sa stiu cum procedez la astfel de exercitii in general . Multumesc :) .

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Fie functia f : R -> R, f(x) = 2x-1. Calculati suma S= f(1) + f(2) +f(3) +.....+f(2012)
Stiu ca numerele cresc din 2 in 2.
f(1)=1
f(2)=3
f(3)=5
f(4)=7 si tot asa.
As vrea sa stiu cum procedez la astfel de exercitii in general . Multumesc :) .