Se consideră funcţia f: R →R, f (x) = 2x -1. Calculaţi f(f(1)) + f(f(2))+ …. f(f(12))

Explicație pas cu pas:

f(1)=2-1=1

f(2)=4-1=3

f(12)=24-1=23

Calculam:

f(1)=1

f(3)=5

f(23)=46-1=45

deci:

1+5+...+45=(4-3)+(8-3)+...+(48-3)=4(1+2+...+12)-12*3=24*13-36=276

Bafta!

[tex]\displaystyle f:\mathbb{R}\to \mathbb{R},\quad f(x) = 2x-1 \\ \\\\ f\Big(f(1)\Big)+ f\Big(f(2)\Big)+...+ f\Big(f(12)\Big) = \\ \\ = \sum\limits_{k=1}^{12}f\Big(f(k)\Big) =\sum\limits_{k=1}^{12}f(2k-1) = \sum\limits_{k=1}^{12}\Big[2(2k-1)-1\Big] = \\ \\ = \sum\limits_{k=1}^{12}(4k-3) = 4\sum\limits_{k=1}^{12}k - \sum\limits_{k=1}^{12}3 = \\ \\\\= 4\cdot \dfrac{12\cdot 13}{2}-3\cdot 12 = 12\cdot 26 - 3\cdot 12 = 12\cdot 23 =\boxed{276}[/tex]