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\(b.\left(1+\frac12\right)\left(1+\frac13\right)\left(1+\frac14\right)\ldots\left(1+\frac{1}{2023}\right)\)
\(=\frac32\cdot\frac43\cdot\frac54\cdot\ldots\cdot\frac{2024}{2023}\)
\(=\frac{3\cdot4\cdot5\cdot\ldots\cdot2024}{2\cdot3\cdot4\cdot\ldots\cdot2023}\)
\(=\frac{2024}{2}=1012\)
\(c.D=\frac{5}{6\cdot37}+\frac{1}{6\cdot43}+\frac{6}{7\cdot43}+\frac{10}{7\cdot59}\)
\(D=7\cdot\left(\frac{5}{37\cdot42}+\frac{1}{42\cdot43}+\frac{6}{43\cdot49}+\frac{10}{49\cdot59}\right)\)
\(D=7\cdot\left(\frac{1}{37}-\frac{1}{42}+\frac{1}{42}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}+\frac{1}{49}-\frac{1}{59}\right)\)
\(D=7\cdot\left(\frac{1}{37}-\frac{1}{59}\right)\)
\(D=7\cdot\frac{22}{2183}\)
\(D=\frac{154}{2183}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2023\cdot2024}\)
\(=\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{2023}-\frac{1}{2024}\)
\(=\frac11-\frac{1}{2024}=\frac{2023}{2024}\)
A=1⋅21+2⋅31+3⋅41+⋯+2023⋅20241
\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \hdots + \frac{1}{2023} - \frac{1}{2024}\)
\(= \frac{1}{1} - \frac{1}{2024} = \frac{2023}{2024}\)
Tổng trên = 1-2^2/2^2 . 1-3^2/3^2 . ..... . 1-100^2/100^2
= -(2^2-1/2^2 . 3^2-1/3^2 . ...... . 100^2-1/100^2 )
= -(1.3/2^2 . 2.4/3^2 . ..... . 99.101/100)
= -(1.2.3. .... .99 . 3.4.5. ... .101 / 2.3.4 . ... . 100 . 2.3.4 . ..... . 100)
= -(1.2.3. ... . 99/2.3.4. .... .100) . (3.4.5. .... .101/2.3.4 . .... . 100)
= -1/100 . 101/2 = -101/200
Tk mk nha
A = (1 + \(\frac12\)).(1 + \(\frac13\)).(1 + \(\frac14\))....(1 + \(\frac{1}{2024}\))
A = \(\frac{2+1}{2}\).\(\frac{3+1}{3}\)+...+\(\frac{2003+1}{2003}\)
A = \(\frac32\).\(\frac43\)...\(\frac{2024}{2003}\)
A = \(\frac{2024}{2}\)
A = 1012
Ta có:
$(1+\dfrac{1}{2})(1+\dfrac{1}{3})(1+\dfrac{1}{4})\cdots(1+\dfrac{1}{2023})$
$= \dfrac{3}{2}\cdot \dfrac{4}{3}\cdot \dfrac{5}{4}\cdots \dfrac{2024}{2023}$
$= \dfrac{3\cdot 4\cdot 5 \cdots 2024}{2\cdot 3\cdot 4 \cdots 2023}$
$= \dfrac{2024}{2}$
$= 1012$
Vậy giá trị bằng $1012$.
Là 1 đó bạn 2024/2 = 1012