Răspuns:
Explicație pas cu pas:
f(x)=3x+1
b)f(0)=3*0+1 ; f(1)=3*1+1 .......f(11)=3*11+1
S2=f(0)+.....+f(11)
S2=(3*0+1)+(3*1+1)+...(3*11+1)->>S=(3*0+3*1+..3*11)+1*12
S2=3(0+1+..+11)+12
S2=3(1+2+3..+11)+12
1+2+3+..+11 ->Suma de tip Gaus
Se stie ca 1+2+3+....+n=[tex]\frac{n*(n+1)}{2}[/tex] , (∀) n∈N*
Deci 1+2+3+...+11= [tex]\frac{11*12}{2}[/tex]
1+2+3+...+11=66
Deci S2=3*66+12
S2=210
a)f([tex](-3)^{0}[/tex])=1+1
f([tex](-3)^{-1}[/tex])=[tex]\frac{1}{3}[/tex]+1
....
f([tex](-3)^{10}[/tex])0[tex]\frac{1}{3^{10} }[/tex]+1
S1=1+[tex]\frac{1}{3}[/tex]+...+[tex]\frac{1}{3^{10} }[/tex]+11*1
S1=11+S3
Pentru a calcula Suma este nevoie de progresiile geometrice
Se stie ca b2=b1*q ->>q=[tex]\frac{1}{3}[/tex]
S3=b1[tex]\frac{q^{11} -1}{q-1}[/tex]
S3=[tex]\frac{ \frac{1}{3} ^{11}-1 }{\frac{1}{3} -1}[/tex]
S1=S3+11....!
