1/2+1/6+1/12+1/20+...+1/2352+1/2450
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}\\ \)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
S=1/2+1/6+...............+1/2450
S=1/1.2+1/2.3+............+1/49.50
S=1-1/2+1/2-1/3+..........+1/49-1/50
S=1-1/50
S=49/50
S=1/2+1/6+...............+1/2450
S=1/1.2+1/2.3+............+1/49.50
S=1-1/2+1/2-1/3+..........+1/49-1/50
S=1-1/50
S=49/50
\(=>S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}+\frac{1}{49.50}\)
\(=>S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
vậy S=49/50
câu 2:
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{2450}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}+\frac{1}{2550}\)
\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{49x50}+\frac{1}{50x51}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}+\frac{1}{50}-\frac{1}{51}\)
\(A=1-\frac{1}{51}=\frac{50}{51}\)
12+1 + 22+2 + 32+3 + 42+4 + ... + 482+48 + 492+49 + 502+50
= (1+2+3+4+..+48+49+50) +(12+22+32+42+...+482+492+502)
Đến đay bạn tự tính nha
A=1/1x2+1/2x3+1/3x4+1/4x5+...+1/49x50+1/50x51
A=2-1/1x2+3-2/2x3+4-3/3x4+...+50-49/49x50+51-50/50x51
A=1-1/2+1/2-1/3+1/3+1/4+...-1/49+1/49-1/50+1/50-1/51
A=1-1/51
A=51/51-1/51
A=50/51
tick nha
1/2+1/6+1/12+1/20+.....+1/2352+1/2450
=1/1.2+1/2.3+1/3.4+1/4.5+.....+1/48.49+1/49.50
=1-/2+1/2-1/31/3-1/4+1/4-1/5+.....+1/48-1/49+1/49-1/50
=1-1/50
=49/50