已知\(n\)为正偶数，用数学归纳法证明\(1- \dfrac {1}{2}+ \dfrac {1}{3}- \dfrac {1}{4}+…+ \dfrac {1}{n+1}=2( \dfrac {1}{n+2}+ \dfrac {1}{n+4}+…+ \dfrac {1}{2n})\)时，若已假设\(n=k(k\geqslant 2)\)为偶数\()\)时命题为真，则还需要用归纳假设再证\(n=(\)　　\()\)时等式成立．