计算:12-22+32-42+52-62+…+n2-（n+1）2=________.（n属于奇数）计算:12-22+32-42+5_数学解答

计算:12-22+32-42+52-62+…+n2-（n+1）2=________.（n属于奇数）
计算:12-22+32-42+52-62+…+n2-（n+1）2=________。（n属于奇数）每个数字后面的2是平方的意思

人气：138 ℃　时间：2019-10-19 16:47:34

解答

1²-2²+3²-4²+……n²-(n-1)²=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]=(-3)+(-7)+……+[-(2n+1)]=-3+[(2n+2)/4-1]×(-4)=-3+[-(2n+2)+4]=-3+(-2n+2)=-2n-1错了，n=3时，1²-2²+3²-4²=-10 -2n-1=-71²-2²+3²-4²+……n²-(n-1)²

=(1-2)(1+2)+(3-4)(3+4)+……+[n-(n+1)][n+(n+1)]
=(-3)+(-7)+……+[-(2n+1)]
=[(2n+2)/4]×(-3)+{[(2n+2)/4][(2n+2)/4-1]×(-4)}/2
=[-3(n+1)/2]-[(n+1)/2](n-1)
=[-(n+1)/2][3+(n-1)]
=[-(n+1)/2](n+2)
=-(n+1)(n+2)/2