Tính tích:
a) M=(1-1/2)x(1-1/3)x(1-1/4)x…x(1-1/n-1)
b) N=(1/2-1)x(1/3-1)x(1/4-1)x…x(1/2003-1)
Cảm ơn mọi người giúp đỡ!
Tính tích:
a) M=(1-1/2)x(1-1/3)x(1-1/4)x…x(1-1/n-1)
b) N=(1/2-1)x(1/3-1)x(1/4-1)x…x(1/2003-1)
Cảm ơn mọi người giúp đỡ!
Đáp án:
$\begin{array}{l}
a)M = \left( {1 – \dfrac{1}{2}} \right).\left( {1 – \dfrac{1}{3}} \right).\left( {1 – \dfrac{1}{4}} \right)…\\
.\left( {1 – \dfrac{1}{{n – 1}}} \right)\\
= \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}….\dfrac{{n – 2}}{{n – 1}}\\
= \dfrac{1}{{n – 1}}\\
b)\\
N = \left( {\dfrac{1}{2} – 1} \right).\left( {\dfrac{1}{3} – 1} \right).\left( {\dfrac{1}{4} – 1} \right)…\left( {\dfrac{1}{{2003}} – 1} \right)\\
= {\left( { – 1} \right)^{2002}}.\left( {1 – \dfrac{1}{2}} \right).\left( {1 – \dfrac{1}{3}} \right)…\left( {1 – \dfrac{1}{{2003}}} \right)\\
= \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}…\dfrac{{2002}}{{2003}}\\
= \dfrac{1}{{2003}}
\end{array}$