Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
Ta có: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}.\)
\(=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{31}+\frac{1}{44}\right)+\left(\frac{1}{61}+\frac{1}{84}+\frac{1}{96}\right)\)
Ta thấy \(\frac{1}{14}< \frac{1}{12}\)
\(\frac{1}{31}< \frac{1}{12}\)
\(\frac{1}{44}< \frac{1}{12}\)
\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}\)
\(=>\frac{1}{14}+\frac{1}{31}+\frac{1}{44}< \frac{1}{12}.3\left(1\right)\)
Ta lại thấy \(\frac{1}{61}< \frac{1}{60}\)
\(\frac{1}{84}< \frac{1}{60}\)
\(\frac{1}{96}< \frac{1}{60}\)
\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}\)
\(=>\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{60}.3\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{5}+3.\left(\frac{1}{12}+\frac{1}{60}\right)\)
\(=>\frac{1}{5}+\frac{1}{14}+\frac{1}{31}+\frac{1}{44}+\frac{1}{61}+\frac{1}{84}+\frac{1}{96}< \frac{1}{2}\)
\(=>Đpcm\)
Ta có :
\(\frac{1}{3^2}< \frac{1}{2\times3};\frac{1}{4^2}< \frac{1}{3\times4};\frac{1}{5^2}< \frac{1}{4\times5};\frac{1}{6^2}< \frac{1}{5\times6};...;\frac{1}{100^2}< \frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{99\times100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow\) \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)
\(B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{8-7}{7.8}\)
\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)
\(B< 1-\frac{1}{8}< 1\left(dpcm\right)\)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2015.2015}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(=1-\frac{1}{2015}=\frac{2014}{2015}< 1\)
=> A < 1 (đpcm)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}< 1-\frac{1}{2017}=\frac{2016}{2017}>\frac{1}{2}\)
\(\Rightarrow\)ko thể cm
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}\)
Ta có :\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}=\frac{1}{2.3}\)
.........
\(\frac{1}{2017^2}=\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{2}-\frac{1}{2017}\)
\(A=\frac{-1}{4}-\frac{1}{2017}=\frac{-2021}{8068}\)
\(\Leftrightarrow A< \frac{1}{2}\) . Vì \(\frac{-2021}{8068}< \frac{1}{2}\)
mình chỉ gợi ý thôi, vì viết cái này mỏi tay lắm thông cảm nha
Ở phần ''a'' bạn hãy đổi ra thành:2=2;4=2;.....sau dó bạn CM \(\frac{1}{2^2}<\frac{1}{1.2}.....\) rồi hãy suy ra nhỏ hơn \(\frac{1}{3}\)
còn phần ''b'' bạn hãy tách ra nha
đặt A=1/2^2+1/3^2+1/4^2+...+1/100^2
B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1-\frac{1}{100}<1\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<1\)
Gọi \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}\) là S
Ta có:
\(S=1+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\right)+\left(\dfrac{1}{8}+\dfrac{1}{9}+...+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{31}\right)\)
\(S< 1+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)\)
\(S< 1+1+1+1+1\)
\(S< 5\)
Vậy \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{31}< 5\)