Toán đại 7

H

hiensau99

[TEX]\frac{1.3}{3.5}+\frac{2.4}{5.7}+..+\frac{(n-1)(n+1)}{(2n-1)(2n+1)}+...+\frac{1005.1007}{2011.2013}[/TEX]

[TEX]=\frac{(3-2).(3+1)}{(2.2-1).(2.2+1)}+\frac{(3-1).(3+1)}{(6-1).(6+1)}+...+\frac{(1006-1).(1006+1)}{(2012-1).(2012+1)}[/TEX]

[TEX]=\frac{1}{4}-\frac{3}{8}.(\frac{1}{2.2-1}-\frac{1}{2.2+1})+\frac{1}{4}-\frac{3}{8}.(\frac{1}{3.2-1}-\frac{1}{3.2+1})+..+\frac{1}{4}-\frac{3}{8}.(\frac{1}{1006.2-1}-\frac{1}{1006.2+1})[/TEX]

[TEX]= 1005.\frac{1}{4}+1005.\frac{-3}{8}+(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-\frac{1}{3.2+1}+...+\frac{1}{1006.2-1}-\frac{1}{1006.2+1})[/TEX]

[TEX]=\frac{1005}{4}+\frac{-3015}{8}+(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013})[/TEX]

[TEX]=\frac{-1005}{8}+(\frac{1}{3}-\frac{1}{2013})[/TEX]

[TEX]=\frac{-1005}{8}+\frac{670}{2013}[/TEX]

[TEX]=\frac{-20177505}{16104}[/TEX]
 
C

computerscience

[TEX]\frac{1.3}{3.5}+\frac{2.4}{5.7}+..+\frac{(n-1)(n+1)}{(2n-1)(2n+1)}+...+\frac{1005.1007}{2011.2013}[/TEX]

[TEX]=\frac{(3-2).(3+1)}{(2.2-1).(2.2+1)}+\frac{(3-1).(3+1)}{(6-1).(6+1)}+...+\frac{(1006-1).(1006+1)}{(2012-1).(2012+1)}[/TEX]

[TEX]=\frac{1}{4}-\frac{3}{8}.(\frac{1}{2.2-1}-\frac{1}{2.2+1})+\frac{1}{4}-\frac{3}{8}.(\frac{1}{3.2-1}-\frac{1}{3.2+1})+..+\frac{1}{4}-\frac{3}{8}.(\frac{1}{1006.2-1}-\frac{1}{1006.2+1})[/TEX]

[TEX]= 1005.\frac{1}{4}+1005.\frac{-3}{8}+(\frac{1}{2.2-1}-\frac{1}{2.2+1}+\frac{1}{3.2-1}-\frac{1}{3.2+1}+...+\frac{1}{1006.2-1}-\frac{1}{1006.2+1})[/TEX]

[TEX]=\frac{1005}{4}+\frac{-3015}{8}+(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013})[/TEX]

[TEX]=\frac{-1005}{8}+(\frac{1}{3}-\frac{1}{2013})[/TEX]

[TEX]=\frac{-1005}{8}+\frac{670}{2013}[/TEX]

[TEX]=\frac{-20177505}{16104}[/TEX]
Các bạn có thể giải thích cho mình hiểu tại sao các số hạng toàn là số dương mà tổng của chúng lại là số âm ?
 
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