2n : (1 + $\frac{1}{1+2}$ + $\frac{1}{1+2+3}$ + $\frac{1}{1+2+3+4}$ +…+ $\frac{1}{1+2+3+4+…+n}$

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1. Đáp án:

    $2(n+1)$

   Giải thích các bước giải:

    $B=2n:\left ( 1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+…+\dfrac{1}{1+2+3+4+…+n} \right )\\
   A=1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+…+\dfrac{1}{1+2+3+4+…+n}\\
   =1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+…+\dfrac{1}{\dfrac{n(n+1)}{2}}\\
   \Rightarrow \dfrac{1}{2}A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+…+\dfrac{1}{n(n+1)}\\
   =\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+…+\dfrac{1}{n(n+1)}\\
   =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+…+\dfrac{1}{n}-\dfrac{1}{n+1}\\
   =1-\dfrac{1}{n+1}\\
   =\dfrac{n+1-1}{n+1}=\dfrac{n}{n+1}\\
   \Rightarrow B=2n:\dfrac{n}{n+1}=2n.\dfrac{n+1}{n}=2(n+1)$

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