(1+1/2)×(1+1/3)×(1+1/4)×.........×(1+1/2005)
LV
Những câu hỏi liên quan
2008+2007/2+2006/3+2005/4+2005/5+........................3/2006+2/2007+1/2008
1/2+1/3+1/4+1/5+....................+1/2009
2008-1/2008=2007/2008
1/2-1/2009=2007/2009
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2008+2007/2+2006/3+2005/4+2005/5+........................3/2006+2/2007+1/2008
1/2+1/3+1/4+1/5+....................+1/2009
Tính
1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2004*2005 + 1/2005*2006
1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2005*2006
= 1- 1/2 + 1/2 - 1/3 + 1/3 -1 /4 + ...+1/2005 - 1/2006
= 1 - 1/2006
= 2005/2006
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Xem thêm câu trả lời
2/2005+1+2^2/2005^2+1+2^3/2005^3+1+....+2^n/2005^n+1....+2^2006/2005^2^2005+1
G=1/1-1/2+1/3-1/4+...+1/2005+1/2006 va H=(1/1+1/2+1/3+....+1/2006)-2.(1/2+1/4+.....+1/2006)
Tính:
S = \(\frac{2}{2005+1}\)+ \(\frac{2^2}{2005^2+1}\)+ \(\frac{2^3}{2005^{2^2}+1}\)+ \(\frac{2^4}{2005^{2^3}+1}\)+ ...+ \(\frac{2^{n+1}}{2005^{2^n}+1}\)+ ...+ \(\frac{2^{2006}}{2005^{2^{2005}}+1}\)
Chứng minh rằng: 2/22+2/32+...+2/20052<1/2+1/3+1/4+...+1/2005
(2008+2007/2+2006/3+2005/4+....+2/2007+1/2008) / (1/2+1/3+1/4+...+1/2009)
Xét tử
2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1
=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008
=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+1/2009)
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Ta có tử số bằng: 2008+2007/2+2006/3+2005/4+…..+2/2007+1/2008
(Phân tích 2008 thành 2008 con số 1 rồi đưa vào các nhóm)
= (1 + 2007/2) + (1 + 2006/3) + (1 + 2005/4) +... + (1 + 2/2007) + ( 1 + 1/2008) + (1)
= 2009/2 + 2009/3 + 2009//4 + ……. + 2009/2007 + 2009/2008 + 2009/2009
= 2009 x (1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009)
Mẫu số: 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009
Vậy A = 2009
Tính: S
S= \(\frac{2}{2005+1}\) + \(\frac{2^2}{2005^2+1}\)+ \(\frac{2^3}{2005^{2^2}+1}\)+ \(\frac{2^4}{2005^{2^3}+1}\)+ .... + \(\frac{2005^{2006}}{2005^{2^{2005}}+1}\)+ .... + \(\frac{2^{n+1}}{2005^{2^n}+1}\)