K=4 /1x4 + 4 /4x7 + 4 /7x11+ .............+4 / 100x 103
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Bạn xem đã viết đúng đề chưa nhỉ. Các thừa số đang cách nhau 3 đơn vị tự nhiên xuất hiện 7 x 11 có 2 thừa số cách nhau 4 đơn vị?
S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\)
S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{4}{7.11}\) - \(\dfrac{1}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\)
S = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{4}{7.11}\) + \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) - \(\dfrac{1}{7.11}\)
S = \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)
S = \(\dfrac{1}{1}\) - \(\dfrac{1}{17}\) - \(\dfrac{1}{77}\)
S = \(\dfrac{16}{17}\) - \(\dfrac{1}{77}\)
S = \(\dfrac{1215}{1309}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{56}-\dfrac{1}{67}\)
\(=1-\dfrac{1}{67}=\dfrac{66}{67}\)
\(E=\dfrac{1}{1\times2}+\dfrac{2}{2\times4}+\dfrac{3}{4\times7}+\dfrac{4}{7\times11}+\dfrac{5}{11\times16}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}\)
\(=1-\dfrac{1}{16}=\dfrac{15}{16}\)
#kễnh
Gọi biểu thức trên là A, ta có:
A = 1/1x2 + 2/2x4 + 3/4x7 + 4/7x11 + 5/11x16
A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/7 + 1/7 - 1/11 + 1/11 - 1/16
A = 1 - 1/16 = 15/16
Đặt \(A=\frac{1}{2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{5}{11\cdot16}\)
\(\Rightarrow A=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(\Rightarrow A=1-\frac{1}{16}\)
\(\Rightarrow A=\frac{15}{16}\)
A=1/1-1/2+1/2-1/4+1/4-1/7+1/7-1/11+1/11-1/16+1/16-1/22+1/22-1/29
A=1/1-1/29
A=28/29