Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bạn quy đồng vs mẫu chung là n(n+1) ta có tử 2 phân số là n+1 và n
=>n+1/n(n+1) - n/n(n+1)=1/n(n+1)
tk mk
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}\)
\(A=\frac{1}{n+1}\)
1)
42n+1+3n+2= (42)n.4 +3n.32
= 16n.4+3n.9
=13n.4+3n.4+3n.9
=13n.4+3n.(4+9)
= 13n.4+3n.13 = 13.(13n-1+3n) chia het cho 13
=> 42n+1+3n+2 chia hết cho 13
2)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
- S = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}\)
- S = \(1-\frac{1}{n+3}\)
\(\Rightarrow\) S < 1 ( đpcm )
=> S = ( 1 -\(\frac{1}{4}\)) + ( \(\frac{1}{4}\)- \(\frac{1}{7}\)) +(\(\frac{1}{7}\)- \(\frac{1}{10}\)) +.....+ (\(\frac{1}{n}\)- \(\frac{1}{n+3}\))
=> S = 1 - \(\frac{1}{4}\)+\(\frac{1}{4}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{10}\)+......+ \(\frac{1}{n}\)- \(\frac{1}{n+3}\)
=> S = 1 - \(\frac{1}{n+3}\)
vậy S = 1- \(\frac{1}{n+3}\)
\(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}\)
\(=\frac{n+1-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\)
\(=\frac{1}{n}-\frac{1}{n+1}\) (đpcm)
Ta có :
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{1}{n.\left(n+1\right)}\)
\(=\frac{1}{n.\left(n+1\right)}\)
Tham khảo nha !!!