\(C=1+1+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)

\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

          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

Đề bài yêu cầu làm j z bn

mình cũng đang định hỏi giống bạn !!!

hảo năm 2018 mới trả lời =)

51/100

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)