[Toán 11]Tính tổng vô hạn

N

noinhobinhyen

$A=1+\dfrac{3}{2}+\dfrac{5}{4}+...+\dfrac{2n+1}{2n}$

$=(n+1)+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2n}$

$=(n+1)+(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{4})+...+(\dfrac{1}{n}-\dfrac{1}{2n})$

$=(n+2)-\dfrac{1}{2n}$
 
You must log in or register to reply here.
Top Bottom