Bài 1 : Tính C= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n-1}{n!}\)
Bài 2 : CMR D=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}< 1\)
Bài 3: Cho biểu thức P=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
a) CMR : P= \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
b) Giải bài toán trên trog trường hợp tổng quát
Bài 4 : CMR: \(\forall n\in Z\left(n\ne0;n\ne1\right)\) thì Q= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\) không phải là số nguyên .
Bài 5 : CMR : S=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{200^2}< \frac{1}{2}\)
1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)
BÀi 1: Thực hiện phép tính ( tính nhanh nếu có thể)
a.\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{71}-\left(-\frac{2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
b.\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
c.\(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{1}{3}-1\frac{1}{15}\right)\)
A= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}\) . cmr a không phải là số nguyên số nguyên
tổng của các phân số cùng mẫu luôn có giá trị của tử thấp hơn giá trị của mẫu => tử không bằng mẫu => A không nguyên
Ta có :
\(A>\frac{1}{4}+\frac{1}{5}+\frac{1}{15}+.....+\frac{1}{15}=\frac{1}{4}+\frac{1}{5}+\frac{10}{15}>1\)
\(A< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+.....+\frac{1}{7}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{9}{7}< 2\)
\(\Rightarrow1< A< 2\)
\(\Rightarrow A\)không phải là số nguyên
Vậy A không phải là số nguyên
Ta có
\(A\)\(>\frac{1}{4}\)\(+\frac{1}{5}+\frac{1}{15}+...+\frac{1}{15}=\frac{1}{4}+\frac{1}{5}+\frac{10}{15}>1\)
B=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}...\cdot\frac{99}{100}\)CMR \(\frac{1}{15}< B< \frac{1}{10}\)
Cho B=\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{99}{100}.\)CMR\(\frac{1}{15}< B< \frac{1}{10}\)
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}\)
CMR: \(\frac{1}{15}< A< \frac{1}{10}\)
\(M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}....\frac{99}{100}\). cmr 1/15 < M <1/10
CMR: \(\frac{1}{5}< \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{6}-\frac{1}{7}< \frac{2}{5}\)
Bài 1:
Chứng minh rằng:
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
Bài 2:
Cho \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}\).
CMR: \(a)A>\frac{4}{3}\); \(b)A< 2,5\)
Đặt : \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta thấy :
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(\frac{1}{7^2}< \frac{1}{6.7}\)
\(.......................\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
Vì \(\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)nên \(\frac{1}{6}< A< \frac{1}{4}\)hay \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
~ Hok tốt ~
Bài 1:
Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)
Ta có:
\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
Ta có:
\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)
\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đ}pcm\right)\)
Bài 2:
\(a)\)Tách tổng A thành ba nhóm:
\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)
\(A>\frac{1}{30}\cdot20+\frac{1}{50}\cdot20+\frac{1}{70}\cdot20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}\)
\(A>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\left(\text{đ}pcm\right)\)
\(b)\)Tách tổng A thành sáu nhóm:
\(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)
\(A< \frac{1}{11}\cdot10+\frac{1}{21}\cdot10+\frac{1}{31}\cdot10+\frac{1}{41}\cdot10+\frac{1}{51}\cdot10+\frac{1}{61}\cdot10\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\left(\text{đ}pcm\right)\)
#Sakura
Bài 1:
ta thấy \(\frac{1}{5^2}< \frac{1}{4.5};\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
.................
=>1/52 +1/62+.....+1/1002<1/(4.5)+1/(5.6)+.....+1/(99+100)
=>1/52 +1/62+.....+1/1002 <1/4 -1/5 +1/5 -1/6 +......+1/99 -1/100
=>1/52 +1/62+.....+1/1002 <1/4 - 1/100 <1/4
CMTT với 1/52 >1/(5.6).......
=>1/52 +1/62+.....+1/1002 >1/5 -1/6 +1/6 - 1/7+ .........+1/100 - 1/101
=>1/52 +1/62+.....+1/1002 >1/5 - 1/101>1/5>1/6
