\(\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{n\times\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)
\(\Rightarrow\frac{n+1-2}{2\left(n+1\right)}=\frac{49}{100}\)
\(\Rightarrow\frac{n-1}{2n+2}=\frac{49}{100}\)
\(\Rightarrow100\left(n-1\right)=49\left(2n+2\right)\)
\(\Rightarrow100n-100=98n+98\)
\(\Rightarrow2n=198\)
=> 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
\(\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{nx\left(n+1\right)}=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{49}{100}\)
\(\frac{1}{n+1}=\frac{1}{100}\)
n + 1 = 100
n = 100 - 1
n = 99
Bn k cho mik nhé!
bạn cài hình ảnh đẹp thế chắc bạn cũng đẹp như thế nhỉ
mình lười 0 giải ra bạn tự giải nhé kết quả là 99
\(\frac{1}{2\times3}\)\(+\)\(\frac{1}{3\times4}+\frac{1}{4\times5}\)\(+...+\frac{1}{n\times\left(n+1\right)}\)\(=\frac{49}{100}\)
\(\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{100}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{49}{100}\)
\(\frac{50}{100}-\frac{49}{100}=\frac{1}{n+1}\)
\(\frac{1}{100}=\frac{1}{n+1}\)
\(n+1=100\)
\(n=99\)
