Với mọi \(x\in Z\) ta có:
\(1+2+3+..+n=\frac{n\left(n+1\right)}{2}\)
=> \(\frac{1}{1+2+3+..+n}=\frac{2}{n\left(n+1\right)}=2\left[\frac{1}{n\left(n+1\right)}\right]=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Có:
\(\frac{1}{1+2}=2\left(\frac{1}{2}-\frac{1}{3}\right)\)
\(\frac{1}{1+2+3}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
.......................................................
\(\frac{1}{1+2+3+4+...+99}=2\left(\frac{1}{99}-\frac{1}{100}\right)\)
Nên:
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+..+99}+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+...+2\left(\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2\cdot\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)
