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Giả sử \(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
\(\Rightarrow100=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}+1+\frac{1}{2}+...+\frac{1}{100}\)
\(\Rightarrow100=1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)+...+\left(\frac{99}{100}+\frac{1}{100}\right)\)
\(\Rightarrow100=1+1+1+...+1\) (100 chữ số 1)
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
Đặt A= 200- (3+\(\frac{2}{3}+\frac{2}{4}+.....+\frac{2}{100}\))
=\(197-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)
=\(\frac{197.2}{2}-\frac{2}{3}-\frac{2}{4}-....-\frac{2}{100}\)
=\(2.\left(\frac{196+1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)
=\(2\left(\frac{196}{2}+\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{100}\right)\)
=\(2\left(98+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-.....-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+.....+1-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.....+\frac{99}{100}\right)\)
Khi đó \(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+....+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=\(\frac{2\left(\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+....+\frac{99}{100}}\)=2(đpcm)
Đặt A là tên biểu thức trên
Ta có: \(A=\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+\frac{2}{5}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}}\)
\(A=\frac{200-2\left(\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)}{\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+....+\left(1-\frac{1}{100}\right)}\)
\(A=\frac{2\left[100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{100}\right)\right]}{100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}\)
\(A=2\)
ta có 200-(3+\(\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\)
=\(1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
=\(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
thay \(2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
ta có \(\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\left(dpcm\right)\)
Ta có :\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
=\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}=\)\(\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\)\(+...+\left(1-\frac{1}{100}\right)\)
=\(\left(1+1+1+....+1\right)\)\(-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(99-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
= \(100-1-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
=\(100-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)= vế trên (đpcm)
\(S=100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(S=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(S=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\RightarrowĐPCM\)
sửa đề câu 1 :
\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{100-1}{100!}\)
\(=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(=1-\frac{1}{100!}< 1\)
sửa đề câu 2
\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+\frac{3.4}{4!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=\left(1+1+\frac{1}{2!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(\Leftrightarrow99-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)
\(\Leftrightarrow1+1+1+...+1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+...+1-\frac{1}{100}\)
\(\Leftrightarrow\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(\Rightarrow100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)\(\left(đpcm\right)\)
1/2+2/3+...+99/100=1-1/2+1-1/3+...+1-1/100=99-(1/2+1/3+,...+1/100)=100-(1+1/2+...+1/100)
Nhận xét: mẫu số của mỗi phân số thuộc số bị trừ trong phép tính trên là số thứ tự của phân số đó trong dãy trên.
Từ đó, ta biết được rằng dãy trên ( số bị trừ có 100 phân số )
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=\left(1+1+1+...+1\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
( Tách 100 thành 100 số 1 )
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}...+\frac{99}{100}\left(đpcm\right).\)
yamate