Question

$Calculati~expresia:\\\\E=E_1+E_2+...+E_n,\\\\unde\\\\E_i=\frac{1}{1*2} +\frac{1}{1*2*3}+...+\frac{1}{1*2*...*(i+1)} \\\\pentru~n=3$

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it E_3=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{1\cdot2\cdot3\cdot4}=\dfrac{^{12)}1}{\ \ 2}+\dfrac{^{4)}1}{\ \ 6}+\dfrac{1}{24}=\dfrac{12+4+1}{24}=\dfrac{17}{24}\\ \\ \\ E_2=\dfrac{1}{1\cdot2}+\dfrac{1}{1\cdot2\cdot3}=\dfrac{^{12)}1}{\ \ 2}+\dfrac{^{4)}1}{\ \ 6}=\dfrac{12+4}{24}=\dfrac{16}{24}\\ \\ \\ E_1=\dfrac{1}{1\cdot2}=\dfrac{^{12)}1}{\ \ 2}=\dfrac{12}{24}\\ \\ \\ E_1+E_2+E_3=\dfrac{12+16+17}{24}=\dfrac{\ 45^{(3}}{24}=\dfrac{15}{8}$