CMR 3/4+8/9+15/16+...+1499/2500>48
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Cmr 3/4 + 8/9 + 15/16 + ... + 2499/2500 > 48
CMR M=3/4+8/9+15/16+..+2499/2500>48
CMR;\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+.....+\frac{2499}{2500}>48\)
Đặt \(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
Chúc bạn học tốt!
Cho c=3/4+8/9+15/16+...+2499/2500
CMR C lớn hơn 48 biết C có 49 số hạng
\(CMR:\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+.....+\frac{2499}{2500}>48\left(49>A>48\right)\)
CTR : 3/4 + 8/9 + 15/16 + 24/25 +....+ 2499/2500 > 48
* Giúp mình với :(
B=3/4+8/9+15/16+...+2499/2500. Chứng minh B>48
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
\(=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow B=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1=48\)
\(\Rightarrow\) B > 48 (đpcm)
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CME:3/4+8/9+15/16+...+2499/2500>48
1.cho A=3/4+8/9+15/16+...+2499/2500
CMR:48<A<49
2.Tìm GTNN của biểu thức sau: E=3x4+ |5y-1| + |5y-2|