Tính tổng : A = c/a1a2+c/a2a3+...+c/an-1an với a2 – a1 = a3 – a2 = … = an – an-1 = k
1 câu trả lời
Ta chứng minh $\dfrac{c}{ab}=\dfrac{c}{k}(\dfrac{1}{a}-\dfrac{1}{b})(b-a=k)$
Vế phải :
$\dfrac{c}{k}(\dfrac{1}{a}-\dfrac{1}{b})\\=\dfrac{c}{ak}-\dfrac{c}{bk}\\=\dfrac{bck - ack}{abk^2}\\=\dfrac{bc-ac}{abk}\\=\dfrac{ck}{abk}=\dfrac{c}{ab}\\\to đpcm$
Vận dụng vào bài ta được :
$C=\dfrac{c}{a_1a_2}+\dfrac{c}{a_2a_3}+...+\dfrac{c}{a_{n-1}a_n}\\=\dfrac{c}{k}(\dfrac{1}{a_1}-\dfrac{1}{a_2}) + \dfrac{c}{k}(\dfrac{1}{a_2}-\dfrac{1}{a_3})+...+\dfrac{c}{k}(\dfrac{1}{a_{n-1}}-\dfrac{1}{a_n})\\=\dfrac{c}{k}(\dfrac{1}{a_1}-\dfrac{1}{a_n})\\=\dfrac{c}{ka_1}-\dfrac{c}{ka_n}$
Vậy $C=\dfrac{c}{ka_1}-\dfrac{c}{ka_n}$
