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a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)
=> \(1-\frac{1}{x+1}=\frac{44}{45}\)
=> \(\frac{x}{x+1}=\frac{44}{45}\)
=> x = 44
b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
.................
\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)
=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)
a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45
=> x=44
b/ 1/22 < 1/1.2; 1/32 < 1/2.3; ....; 1/452 < 1/44.45
=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1
=> A < 1
1/2^2+1/3^3+.....+1/45^2 < 1/1.2+1/2.3+...+1/44.45=1-1/2+1/2-1/3+...+1/44-1/45=1-1/45=44/45 <1
Suy ra : 1/2^2 +...+1/45^2<1
\(\text{Ta có: }\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{45.45}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{44.45}\)
\(=1-\frac{1}{45}=\frac{44}{45}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{44}{45}< 1\)
TA CÓ:
A=30+3+32+33+........+311
(30+3+32+33)+....+(38+39+310+311)
3(0+1+3+32)+......+38(0+1+3+32)
3.13+....+38.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)
a) S = 2 + 22 + 23 + 24 +.....+ 29 + 210
= (2 + 22) + (23 + 24) +.....+ (29 + 210)
= 2(1 + 2) + 23(1 + 2) +....+ 29(1 + 2)
= 3.(2 + 23 +.... + 29) chia hết cho 3
=> S = 2 + 22 + 23 + 24 +.....+ 29 + 210 chia hết cho 3 (Đpcm)
b) 1+32+33+34+...+399
=(1+3+32+33)+....+(396+397+398+399)
=40+.........+396.40
=40.(1+....+396) chia hết cho 40 (đpcm)
ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{45^2}< \frac{1}{44.45}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}...+\frac{1}{44.45}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{44}-\frac{1}{45}\)
\(=1-\frac{1}{45}< 1\) (1)
mà \(\frac{1}{2^2}>0;\frac{1}{3^2}>0;\frac{1}{4^2}>0;...;\frac{1}{45^2}>0\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}>0\)(2)
Từ (1);(2) \(\Rightarrow0< M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{45^2}< 1\)
=> M không phải là số tự nhiên ( đ p c m)