C=3/4*8/9*15/16...2499/2500
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\(C=1+1+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
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
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
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(=\left(1+1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
\(=49.1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
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)
\(C=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{2499}{50^2}=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot50\cdot50}=\frac{1\cdot51}{2\cdot50}=\frac{51}{100}\)
\(C=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}\)
\(C=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+...+\frac{2499}{50^2}\)có 49 số hạng
Bài này là bài chứng minh mà bạn