Cho M = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....\frac{1}{18.19.20}\)
Chứng minh M<\(\frac{1}{4}\)
Mình đang cần rất gấp! Bạn nào giải nhanh mình sẽ tick cho! ^.^ ^-^ ^.^ ^-^
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Cho \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}.\)
Chứng minh rằng:\(B< \frac{1}{4}.\)
2B=\(\frac{2}{1.2.3}\)+.....+\(\frac{2}{18.19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\).......+\(\frac{1}{18.19}\)-\(\frac{1}{19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{19.20}\)
B=\(\frac{1}{1.2}\):2-\(\frac{1}{19.20}\):2
B=\(\frac{1}{1.2}\).\(\frac{1}{2}\)-\(\frac{1}{19.20}\).\(\frac{1}{2}\)
=\(\frac{1}{4}\)-\(\frac{1}{19.20.2}\)<\(\frac{1}{4}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{19.20}\)
\(B=\left(\frac{1}{2}-\frac{1}{19.20}\right):2\)
\(B=\frac{189}{760}\)
A=1/2{(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+...+(1/18*19-1/19*20)}
=1/2{1/1*2-1/19*20}
=1/2*189/380
=189/760
vì 189/760<1/4
nên A=...<1/4
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}<\)\(\frac{1}{2}\)
\(2A<\)\(\frac{1}{2}\)
\(\Rightarrow A<\)\(\frac{1}{4}\)
Vậy \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}<\)\(\frac{1}{4}\)
$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{19.20}\right)=\frac{1}{4}-\frac{1}{2.19.20}
B=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}< 3\)
=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+............+\frac{1}{18.19.20}\)
=\(\frac{2}{1.2.3.2}+\frac{2}{2.3.4.2}+............+\frac{2}{18.19.20.2}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}............+\frac{1}{18.19}-\frac{1}{19.20}\)
=\(\frac{1}{1.2}-\frac{1}{19.20}\)
=\(\frac{189}{380}\)
A = 5/20.22 + 5/22.24+...+5/79.81
A = 5/2 . (2/20.22 + 2/22.24 + ... + 2/79.81)
A = 5/2 . (1/20 - 1/22 + 1/22 - 1/24 + ... + 1/79 - 1/81)
A = 5/2 . (1/20 - 1/81)
A = 5/2 . 61/1620
A = 61/648
B = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.29
2B = 2/1.2.3 + 2/2.3.4 + ... + 2/18.19.20
\(\Rightarrow\)B = 1/1.2 + 1/2.3 + ... + 1/19.20
\(\Rightarrow\)B = 1/1.2 - 1/19.20
B = 1/2 - 1/380
B = 189/380
bạn ơi số cuối sau \(\frac{12}{18.19.20}\) là zì dậy bị che mất rồi mk k thấy
A=1/2(1/1.2-1/2.3+1/2.3-1/3.4+...+1/18.19-1/19.20
A=1/2(1/1.2-1/19.20)
A=1/2(1/2-1/380)
A=1/2.189/380
A=189/760
Mà 189/760<1/4
=>A<1/4
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{18.19.20}\)
2A = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{18.19.20}\)
2A = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{18.19}-\frac{1}{19.20}\)
2A = \(\frac{1}{1.2}-\frac{1}{19.20}\)
2A = \(\frac{189}{360}\)
A = \(\frac{189}{360}:2\)
Vậy A = \(\frac{189}{760}<\frac{189}{756}=\frac{1}{4}\)
k nha?