làm thế nào

= 1.

Cách giải: Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)

\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)

\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)

\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)

\(=\frac{49}{50}+\frac{1}{50}=1\)

Ket qua la 1 con neu muon xem cach giai thi vao chtt

Đặt \(B=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}\)

\(B=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{4950}\)

\(\Rightarrow B=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow B=2\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(\Rightarrow B=2.\frac{49}{100}\)

\(\Rightarrow B=\frac{49}{50}\)

\(\Rightarrow A=B+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)

1/1+2+1/1+2+3+1/1+2+3+4+...+1/1+2+3+...+99+1/50

=1/(2+1).2:2+1/(3+1).3:2+1/(4+1).4:2+...+1/(99+1).99:2+1/50

=2/2.3+2/3.4+2/4.5+...+2/99.100+1/50

=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)+1/50

=2.49/100+1/50=49/50+1/50=1

tick nha ^^

xin lõi bạn, tui không biết giải

lm` di mik tick

\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)

\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)

\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)

\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)

Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T

JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)