chứng tỏ \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + .........+ \(\frac{1}{100.100}\) < 1
Giup mk tich cho
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+...........+\(\frac{1}{100.100}\)so sanh voi 1
Đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}\)
Mà \(\frac{99}{100}< 1\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\)
Tính:
\(C=\) \(\frac{1.1!}{1!.2!}+\frac{2.2!}{2!.3!}+\frac{3.3!}{3!.4!}+......+\frac{100.100!}{100!.101!}\)
toán đúng rồi đó ban, nhưng mình làm rồi
Cho A= ( \(\frac{1}{2.2}\)-1).( \(\frac{1}{3.3}\)-1).( \(\frac{1}{4.4}\)-1)...( \(\frac{1}{100.100}\)-1)
So sánh A với -\(\frac{1}{2}\)
\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{1010.1010}\)< 1. Chúng tỏ tổng này nhỏ hơn 1
Chứng minh rằng : 1/2.2+1/3.3+...+1/100.100<1
Có : 1/2^2+1/3^2+....+1/100^2 < 1/1.2+1/2.3+....+1/99.100 = 1-1/2+1/2-1/3+....+1/99-1/100 = 1-1/100 < 1
=> ĐPCM
k mk nha
Chứng minh : 1/2.2+1/3.3+......+ 1/100.100<1
Ta có : 1/2.2 < 1/1.2
1/3.3 < 1/2.3
.
.
.
1/100.100<1/99.100
==> 1/2.2+1/3.3+...+1/100.100 < 1/1.2 + 1/2.3+....+1/99.100
=> A < 1-1/100
=> A<99/100<100/100=1
==> a<1
Chứng minh rằng:
a) A= 1/ 2.2 + 1/3.3 + 1/4.4 +....+ 1/100.100 < 1
1/2.2+1/3.3+1/4.4+....+1/100.100<1
1/2.2 < 1/1.2
1/3.3 < 1/2.3
..................
1/100.100 < 1/99.100
=> <
Ta có: \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+....+\frac{1}{100.100}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
\(\frac{1}{4^2}<\frac{1}{3.4}\)
.....
\(\frac{1}{100^2}<\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}<1\left(đpcm\right)\)
1/2.2 < 1/1.2
1/3.3 < 1/2.3
..................
1/100.100 < 1/99.100
=> <
S=\(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+...+\frac{1}{49.49}+\frac{1}{50.50}\)=?
S= 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4 +...+ 1/50 - 1/50
S= 0 + 0 + 0 +...+ 0
S= 0
\(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{49.49}+\frac{1}{50.50}\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{50}-\frac{1}{50}\)
\(=0+0+...+0\)
\(=0\)