Question

Fie f:R-R f(x) =x+5. Sa se calculeaze.
$f(2) + f(2 {}^{2} ) + .... + f(2 {}^{7})$
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Answer 1

 

$\displaystyle\bf\\f:R\to R,~f()x=x+5\\\\f(2)+f(2^2)+...+f(2^7)=\\\\=(2^1+5)+(2^2+5)+...+(2^7+5)=~~~(Avem~7~termeni.)\\\\=\underbrace{5+5+...+5}_{7~termeni}+2^2+2^2+...+2^7=\\\\\\=7\times5+2^{7+1}-1-1=\\\\35+2^8-2=\\\\35+256-2=\boxed{\bf289}$