cho H =2+3/4+8/9+15/16+...+2499+2500
chứng minh: H> 50
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
KT
1
LB
4 tháng 5 2018
\(H=2+\dfrac{4-1}{4}+\dfrac{9-1}{9}+\dfrac{16-1}{16}+..+\dfrac{2500-1}{2500}\)\(H=2+49-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\dfrac{1}{4}-\dfrac{1}{9}-\dfrac{1}{16}-..-\dfrac{1}{2500}\)
\(H-51=-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)\)
\(-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..+\dfrac{1}{50.50}\right)>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}\right)\)
\(H-51>-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+..+\dfrac{50-49}{49.50}\right)\)
\(H-51>-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(H-51>-\left(1-\dfrac{1}{50}\right)\)
\(H>-\dfrac{49}{50}+51>50\)
26 tháng 3 2018
\(A=1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+\frac{1}{2500}\)
\(A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+\frac{1}{50^2}=\left(1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}
{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)\)(từ 2 đến 50 có 49 số nên có 49 số 1)
\(A=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)<49\) (1)
Nhận xét: \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{50^2}<\frac{1}{49.50}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-
\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}<1\)=> \(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}
{4^2}...+\frac{1}{50^2}\right)>-1\)
=> \(A=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...+\frac{1}{50^2}\right)>49-1=48\)(2)
từ (1)(2) => 48 < A < 49 => A không là số tự nhiên.
Bạn lên mạng có đấy
AK
26 tháng 3 2018
CM : 3/4 + 8/9 + 15/16 + ...+ 2499/2500 > 48 => 2 + 3/4 + 8/9 + 15/16 + ...+ 2499/2500 > 50 hay H > 50
Tham khảo tại : https://olm.vn/hoi-dap/question/88888.html
Chúc học tốt !!!
NM
1
XO
27 tháng 8 2020
\(C=1+1+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
51 số hạng 49 số hạng
= \(51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)\)
\(>51-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\right)=51-\left(\frac{1}{2}-\frac{1}{51}\right)=51-\frac{1}{2}+\frac{1}{51}\)
\(=50,5+\frac{1}{51}>50\left(đpcm\right)\)
Vậy C > 50
ND
1
LB
14 tháng 5 2018
DẶT A= BIỂU THỨC TRÊN
A=2+1+1+..+1-(1/4+1/9+...+1/2500)
ĐẶT S=1/4+1/9+...+1/2500
S=1/2^2+1/3^2+...+1/50^2
SÓ SỐ HẠNG CỦA S:
(50-2)/1+1=49
SUY RA
1+1+...+1=49
SUY RA A=2+49-S
A=51-S
TAO CÓ :
S<1/1.2+1/2.3+...+1/49.100
S<1-1/2+1/2-1/3+...+1/49-1/50
S<1-1/50
S<49/50
SUY RA A>51-49/50
SUY RA A>50
DT
1
8 tháng 5 2016
\(A=2+\frac{3}{4}+\frac{8}{9}+......+\frac{2499}{2500}\)
\(A=2+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+.....+\left(1-\frac{1}{2500}\right)\)
\(A=2+1-\frac{1}{4}+1-\frac{1}{9}+.........+1-\frac{1}{2500}\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+....+\frac{1}{2500}\right)\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{50^2}\right)\)
Vì mỗi số 1 đều đi với 1 phân số nên có số số 1 là: (50-1)/1+1=50(số)
\(A=52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{50^2}\right)\)
\(\frac{1}{2^2}<\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}\)
.........
\(\frac{1}{50^2}<\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{49}{50}\)
\(\Rightarrow52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)>52-\frac{49}{50}\)
\(\Rightarrow A>51\frac{1}{50}\)
Vì\(51\frac{1}{50}>50\Rightarrow A>50\)
Sai đề à bạn H>49 thì đc
\(H=2+\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(=2+1-\frac{1}{4}+1-\frac{1}{9}+1-\frac{1}{16}+...+1-\frac{1}{2500}\)
\(=2+49-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(=51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)\)
Do \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{50.50}< \frac{1}{49.50}\)
Nên \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< 1\)
\(\Rightarrow H=51-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}\right)>51-1=50\)
Vậy H>50