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=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100
1/2-1/a+1=49/100
1/a+1 = 1/2-49/100
1/a+1=1/100
a+1=100
a=99
=1/2-1/3+1/3-1/4+.......+1/a-1/a+1=49/100
1/2-1/a+1=49/100
1/a+1 = 1/2-49/100
1/a+1=1/100
a+1=100
a=99
\(\frac{1}{2\times3}+\frac{1}{3\times4}+............+\frac{1}{a\times\left(a+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..............+\frac{1}{a}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{a+1}=\frac{1}{2}-\frac{49}{100}\)
\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\)
\(\Rightarrow a+1=100\)
\(\Rightarrow a=99\)
Đáp số là a = 99 nha còn cách làm thì Nguyễn Hung Phat đã làm rồi nha
T ik nha bạn =))
Chúc bạn học tốt nhé !!!
1/2x3 +1/3x4 +..........+ 1/ax(a+1)=299/600
=>1/2-1/3+1/3-1/4+.........+ 1/a -1/a+1=299/600
=>1/2-1/a+1=299/600
=>a-1/2a=299/600
=>a=300
\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)
\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)
\(\Rightarrow100n-100=98n+98\)
\(\Rightarrow2n=198\)
=> n = 99
Vậy n = 99
\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)
\(\frac{1}{2}\)-\(\frac{1}{n+1}\)=\(\frac{49}{100}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)-\(\frac{49}{100}\)
\(\frac{1}{n+1}\)=\(\frac{1}{100}\)
=> n+1=100
n=100-1
n=99
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{2}-\frac{1}{99}\)
\(=\frac{99}{198}-\frac{2}{198}\)
\(=\frac{97}{198}\)
\(\frac{A}{198}=\frac{97}{198}=>A=198x97:198=97\)
Dạ thưa Vũ a = 99