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Gọi tổng đã cho là A.
Ta có: A = 1/1x2x3 + 1/2x3x4 + 1/3x4x5 +1/4x5x6 .... + 1/10x11x12 Chú ý rằng: 1/1x2x3 = 1/2 x(1/1x2 - 1/2x3) 1/2x3x4 =1/2 x (1/2x3 - 1/3x4) 1/3x4x5 = 1/2 x (1/3x4 - 1/4x5) ..........
Từ đó suy ra: A = 1/2 x (1/1x2 - 1/2x3 + 1/2x3 - 1/3x4+1/3x4 - 1/4x5+...+ 1/10x11 - 1/11x12) = 1/2 x (1/1x2 - 1/11x12) =1/2 x ( 1/2 - 1/132) = 65/264
Gọi tổng đã cho là A. Ta có:
A = 1/1x2x3 + 1/2x3x4 + 1/3x4x5 +1/4x5x6 .... + 1/10x11x12
Chú ý rằng:
1/1x2x3 = 1/2 x(1/1x2 - 1/2x3)
1/2x3x4 =1/2 x (1/2x3 - 1/3x4)
1/3x4x5 = 1/2 x (1/3x4 - 1/4x5)
..........
Từ đó suy ra:
A = 1/2 x (1/1x2 - 1/2x3 + 1/2x3 - 1/3x4+1/3x4 - 1/4x5+...+ 1/10x11 - 1/11x12)
= 1/2 x (1/1x2 - 1/11x12)
=1/2 x ( 1/2 - 1/132)
= 65/264
A = 1/4 + ... +1/84
A = 2/8 + 2/24 + ... + 2/168
A = 2/2.4 + 2/4.6 + ... + 2/12.14
A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14
A = 1/2 - 1/14
A = 6/14 = 3/7
A = 2/8 + 2/24 + ... + 2/168
A = 2/2.4 + 2/4.6 + ... + 2/12.14
A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14
A = 1/2 - 1/14
A = 3/7
K = \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}\)
\(=\frac{1}{2}\times\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{8}\right)\)
\(=\frac{1}{2}\times\frac{7}{8}=\frac{7}{16}\)
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)NHÂN CẢ TỬ VÀ MẪU CỦA TỪNG P/S VỚI 2 TA ĐƯỢC:
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{12.14}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)
Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)
\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)
\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)