填空题数学归纳法证明（n+1）+（n+2）+…+（n+n）=的第二步中，当n=k+1时等式左边与n=k时等式左边的差等于________．
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3k+2解析分析：根据条件分别求出n=k时左边的式子：（k+1）+（k+2）+…+（k+k），n=k+1时左边的式子：左边=（k+1+1）+（k+1+2）+…+（k+1+k-1）+（k+1+k）+（k+1+k+1），从而可求出当n=k+1时等式左边与n=k时等式左边的差．解答：由题意，n=k时，则（k+1）+（k+2）+…+（k+k）=当n=k+1时，左边=（k+1+1）+（k+1+2）+…+（k+1+k-1）+（k+1+k）+（k+1+k+1）=（k+2）+（k+3）+…+（k+k）+（k+1+k）+（k+1+k+1）=（k+1）+（k+2）+（k+3）+…+（k+k）+（k+1+k）+（k+1+k+1）-（k+1）=+3k+2∴当n=k+1时等式左边与n=k时等式左边的差等于3k+2故