Question

Se considera f:R=>R. F(x) =x+1.
Calculați suma f(2)+f(2^2)+...f(2^2018)
^=putere

Answer 1

$f:\mathbb_R \rightarrow\mathbb_R , \quad f(x) = x+1\\ \\ S= f(2)+f(2^2)+...f(2^{2018})\\ S = \sum\limits_{k=1}^{2018} f(2^k) \\S =\sum\limits_{k=1}^{2018}(2^k + 1) \\ S =\sum\limits_{k=1}^{2018}(2^k) + 2018 \\ S-2018 = \sum\limits_{k=1}^{2018}2^k\Big|\cdot 2\\ 2S-4036 = \sum\limits_{k=1}^{2018}2^{k+1} \\ 2S-4036 = \sum\limits_{k=2}^{2019}2^k\\ 2S-4036 = 2^1 + \sum\limits_{k=2}^{2019}(2^k)-2^1 \\ 2S - 4036 = \sum\limits_{k=1}^{2019}(2^k)-2 \\ 2S - 4036 = \sum\limits_{k=1}^{2018}(2^k) + 2^{2019} +2018-2018- 2 \\ 2S - 4036 = \sum\limits_{k=1}^{2018}(2^k) + 2018 + 2^{2019} -2018- 2 \\ 2S - 4036 = \sum\limits_{k=1}^{2018}(2^k+1)+ 2^{2019}-2020 \\ \\ 2S -4036+2020 = S +2^{2019}\\\\ 2S-S = 2016+2^{2019}\\\\ S = 2016 + 2^{2019}$

Answer 2

f(2)=f(2^1)=2+1
f(2²)=2²+1
f(2³)=2³+1
................
f(2^2018) =2*2018+1
suma =2+2²+2³+...2^2018+2018
dar 2+2²+..2^2018=1+2+2²+...+2^2018-1=(2^2019-1)/(2-1) -1=2^2019-2
atunci toatr suma este
2^2019-2+2018=2^2019+2016
as simple as that!!