chứng minh (quy nạp)

M

minhtuyb

Gợi ý:
Đặt $S_n =\dfrac{1}{n+1}+...+\dfrac{1}{3n+1}$
Với $n=1: S_1>1$ (đúng)
Giả sử $S_k>1$, ta cần c/m: $S_{k+1}>1:$
$$S_{k+1} =(\dfrac{1}{n+1}+...+\dfrac{1}{3n+1})-\dfrac{1}{n+1}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}\\ >1-\dfrac{1}{n+1}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}$$
Vậy ta cần c/m $-\dfrac{1}{n+1}+\dfrac{1}{3n+2}+\dfrac{1}{3n+3}+\dfrac{1}{3n+4}>0$
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