Question

1²-2²+3²-4²+5²-6²+…-100²+101²等于多少？急用,,,,,

Answer 1

1²-2²+3²-4²+5²-6²+…-100²+101²

=101²-100²+99²-98²……+3²-2²+1²
=(101+100)(101-100)+(99+98)(99-98)……+(3+2)(3-2)+1
=201+197+……+5+1
等差数列，公差 d=4，项数 n=（201-1)/4=50
原式=(201+1)x50/2=5050

Answer 2

5151

Answer 3

原式=—(2^2-1^2+4^2-3^2+5^2-6^2+...+101^2-99^2-101^2)
=-[(2+1)*(2-1)+(4+3)*(4-3)+...+(100+99)*(100-99)-101^2]
=-(1+2+3+4+...+100-101^2)
=5151
前边用的是平方差公式，后边累加求和这个应该初中时候学过吧
还有不清楚之处可追问

Answer 4

1²-2²+3²-4²+5²-6²+…-100²+101²
=1²+(-2²+3²-4²+5²-6²+…-100²+101²)
=1²+[(-2²+3²)+(-4²+5²)....+(-100²+101²)]
=1²+[(-2+3)(-2-3)+(-4+5)(-4-5)....+(-100+101)(-100-101)]
=1²+[(-1)(-2-3)+(-1)(-4-5).....+(-1)(-100-101)]
=1²+[(-1)(-2-3-4-5-...-100-101)]
=1²+(2+3+4+5+...+100+101)
=1+2+3+4+5+...+101
=5151

为了lz看懂全部手打，没有跳步骤，望采纳

Answer 5

1²-2²+3²-4²+5²-6²+…-100²+101²
=(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100)+101^2
=-[3+7+11+...+199]++101^2
=-[3+199]*50/2+101^2
=-202*50/2+101^2
=-101*50+101^2
=-101(50-101)
=-101*(-51)
= 5151

Answer 6

1²-2²+3²-4²+5²-6²+…-100²+101²

=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+.........+(99-100)(99+100)+101²
=-1-2-3-4-5-6-.........-99-100+101²
=-(1+100)*100/2+101²
=101²-101*50
=5151