Question

E(1)+E(1/2)+E(1/3)+...+E(1/n)=?
si o explicatie mica

Answer 1

$E(x)=\frac{1}{x}+1 =\ \textgreater \ E(x)=\frac{x+1}{x},x \neq 0$
$E(1)=\frac{1+1}{1}=2, E(\frac{1}{2})=\frac{\frac{1}{2}+1}{\frac{1}{2}}=\frac{3}{2}*2 = 3, \\ E(\frac{1}{3})=\frac{\frac{1}{3}+1}{\frac{1}{3}} = \frac{4}{3}*3=4$
Deducem ca E(1/n)=n+1
Suma noastra este:
E(1)+E(1/2)+E(1/3)+...+E(1/n)=2+3+4+...+(n+1)=(1+1)+(2+1)+(3+1)+...+(n+1)=(1+2+3+...+n)+n=$\frac{n(n+1)}{2}+n=\frac{n^2+n+2n}{2}=\frac{n^2+3n}{2}=\frac{n(n+3)}{2}$
In suma noastra am scris toti termenii ca ultimul si i-am grupat si am facut calculul.
Succes