Giup mk tich cho
\(\frac{1}{2.2}\)+\(\frac{1}{3.3}\)+...........+\(\frac{1}{100.100}\)so sanh voi 1
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 -\(\frac{1}{2}\)
chứng tỏ \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + .........+ \(\frac{1}{100.100}\) < 1
Ta có : \(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4.4}< \frac{1}{3.4}\)
...................
\(\frac{1}{100.100}< \frac{1}{99.100}\)
Suy Ra : \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+......+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\)
\(\frac{1}{2.2}+\frac{1}{3.3}+.....+\frac{1}{100.100}< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Ta có : \(\frac{1}{2.2}\)\(< \frac{1}{1.2}\)
\(\frac{1}{3.3}\)\(< \frac{1}{2.3}\)
\(\frac{1}{4.4}\)\(< \frac{1}{3.4}\)
...... .... ......
\(\frac{1}{100.100}\)\(< \frac{1}{99.100}\)
\(\Rightarrow\)\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ \(\frac{1}{4.4}\)+ ..... + \(\frac{1}{100.100}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ..... + \(\frac{1}{99.100}\)
\(\frac{1}{2.2}\)+ \(\frac{1}{3.3}\)+ .... + \(\frac{1}{100.100}\)< \(1-\frac{1}{100}=\frac{99}{100}< 1\)
1/2.2 < 1/1.2
1/3.3 < 1/2.3
1/4.4 < 1/3.4
1/100.100 < 1/ 99.100
Nên 1/2.2 + 1/3.3 +1/4.4 + .... +1/100.100 < 1/1.1 +1/2.3+1/3.4 +......+ 1/99.100
1/2.2 + 1/3.3+.... 1/100.100 < 1 - 1/100 = 99/100 < 1
ta còn có 1 cách làm ngắn gọn hơn
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
So sanh
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{90}+\frac{1}{100}.\)va 100
cac ban giai chi tiet giup mk voi
Bạn sai đè thì phải,đúng phải là 1/99
Ta thấy:Từ 1->1/100 có 100 số.
Ta có:100=1.100
Vì 1=1 ;1/2<1 ;1/3<1 ;1/4<1 ;... ;1/90<1 ;1/100<1.
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 1.100=100\)
\(\Rightarrow1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}< 100\)
1:cho A=\(\frac{1}{7}\)+\(\frac{1}{13}\)+\(\frac{1}{25}\)+\(\frac{1}{49}\)+\(\frac{1}{97}\)
hay so sanh tong A voi \(\frac{1}{3}\)
giup minh nha.
minh cho 3 tich.
lười làm quá đúng cho tui đi rồi tui làm
Đáp án là : A<1/3
Bạn quy đồng mẫu rồi cộng lại sau đó so sánh A với 1/3. Nếu A và 1/3 ko cùng mẫu hoặc tử thì quy đồng mẫu hoặc quy đồng tử rồi so sánh. Dạng này dễ mà sao bây giờ bạn vẫn còn hỏi?
Chúc bạn học tốt và chăm hơn nhé.
1/2.2 + 1/3.3 + 1/4.4 +....+ 1/99.99 + 1/100.100
giúp mk nha
1/2.2 + 1/3.3 + 1/4.4 +....+ 1/99.99 + 1/100.100
= 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/98.99 + 1/99.100
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100
= 1/1 - 1/100
= 99/100
So sánh A và 1 :
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}\)
Ta có:\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}< 1\)
=>A<1
\(\text{Ta có: }\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};.....;\frac{1}{10.10}< \frac{1}{9.10}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(\Rightarrow A< 1-\frac{1}{10}\)
\(\Rightarrow A< \frac{9}{10}< 1\)
A = 1/2.2 + 1/3.3 + 1/4.4 + ... + 1/10.10
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10
A < 1 - 1/10 < 1
=> A < 1
1:cho A=\(\frac{1}{7}\) +\(\frac{1}{13}\) +\(\frac{1.}{25}\) +\(\frac{1}{49}\) +\(\frac{1}{97}\)
hay so sanh tong A voi 13
giup minh nha.
minh cho 3 tich.
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
=> <