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Bình chọn tăng 7 Bình chọn giảm

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Chứng minh rằng với mọi số tự nhiên $n \neq 0$ ta đều có :

$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{(3n-1).(3n+2)}=\frac{n}{6n+4}$

Sửa 31-03-16 07:19 PM

Lionel Messi
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Hỏi 31-03-16 06:29 PM

Hàn Thiên Dii
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1 Đáp án

Thời gian Bình chọn

Bình chọn tăng 2 Bình chọn giảm

$VT=\frac{1}{3}(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{(3n-1)(3n+2)})$

$=\frac{1}{3}.(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2})$

$=\frac{1}{3}(\frac{1}{2}-\frac{1}{3n+2})=\frac{1}{3}.\frac{3n}{6n+4}=\frac{n}{6n+4}$

Trả lời 31-03-16 07:24 PM

Lionel Messi
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