tính tích

  • Thread starter nguyntho353@yahoo.com.vn
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toiyeu9a3

$\dfrac{1}{n^2 - 1} = \dfrac{1}{(n - 1)(n + 1)}$ = \dfrac{1}{2}(\dfrac{1}{n - 1} - \dfrac{1}{n + 1})$
Áp dụng ta có:
$C = \dfrac{1}{2}( 1 - \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{5} + ...+ \dfrac{1}{97} - \dfrac{1}{99} + \dfrac{1}{98} - \dfrac{1}{100})$
$= \dfrac{1}{2}(1 - \dfrac{1}{99} + \dfrac{1}{2} - \dfrac{1}{100})$
yêu cầu sửa latex
 
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