Question

срочно надо хелп плиз​

Answer 1

Воспользуемся свойством:

$\boxed{\frac{1}{a(a+1)} =\frac{1}{a} -\frac{1}{a+1}}$

$\frac{1}{1*2} +\frac{1}{2*3}+\frac{1}{3*4}+...+\frac{1}{98*99} +\frac{1}{99*100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+... \\ \\ ...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100} =0.99$

Ответ: 0,99

Answer 2

Ответ:  0,99 .

Пошаговое объяснение:

$\boxed {\frac{1}{n\cdot (n+1)}=\frac{1+n-n}{n\cdot (n+1)}=\frac{n+1}{n\cdot (n+1)}-\frac{n}{n\cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1}}\\\\\\\frac{1}{1\cdot 2}+\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+...+\frac{1}{97\cdot 98}+\frac{1}{98\cdot 99}+\frac{1}{99\cdot 100}=\\\\=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{97}-\frac{1}{98})+(\frac{1}{98}-\frac{1}{99})+(\frac{1}{99}-\frac{1}{100})=\\\\=1-\frac{1}{100}=\frac{99}{100}=0,99$