Question

数学题，求高手解答，谢谢。（要求是简便方法算）

1+2*3+4*5+6*7+8*9+…………+94*95+96*97+98*99+100等于多少？（*代表乘法符号）。要简便方法过程和结果。谢谢

Answer 1

原式=﹙1＋2＋4＋6＋8.....＋98＋100﹚﹢﹙2²＋4²﹢6²﹢......98²)
=[1+(2+100)×50÷2]+[(1×2﹚²﹢﹙2×2﹚²﹢﹙2×3﹚²＋........﹢﹙2×49﹚²]
=2551+2²(1²﹢2²﹢3²﹢.......﹢49²﹚
=2551+4×49×50×99÷6
=164251

Answer 2

首先要知道1²﹢2²÷3²÷.........﹢n²=n(n+1)(2n+1)/6
原式=﹙1＋2＋4＋6＋8.....＋98＋100﹚﹢﹙2²＋4²﹢6²﹢......98²)
=[1+(2+100)×50÷2]+[(1×2﹚²﹢﹙2×2﹚²﹢﹙2×3﹚²＋........﹢﹙2×49﹚²]
=2551+2²(1²﹢2²﹢3²﹢.......﹢49²﹚
=2551+4×49×50×99÷6
=164251

Answer 3

原式=1+2*(2+1)+4*(4+1)+......+98*(98+1)+100=1+﹙2²＋4²﹢6²﹢......98²)
+（2＋4＋6＋8.....＋98）＋100

Answer 4

1+2*3+4*5+6*7+8*9+…………+94*95+96*97+98*99+100
=1+2^2+2+4^2+4+6^2+6+8^2+8+......+98^2+98+100
=1+2+4+6+8++......+98+100+2^2+4^2+6^2+8^2+......+98^2
=1+2*(1+2+3+4+......+49+50)+2^2*(1^2+2^2+3^2+4^2+......+49^2)
=1+2*51*25+4*49*50*99/6
=1+50*51+100*49*33
=1+2550+100*(50-1)*33
=2551+165000-3300
=164251

注:
1+2+3+......+n
=n(n+1)/2

n^2=n(n+1)-n
1^2+2^2+3^2+......+n^2
=1*2-1+2*3-2+....+n(n+1)-n
=1*2+2*3+...+n(n+1)-(1+2+...+n)

由于n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
所以1*2+2*3+...+n(n+1)
=[1*2*3-0+2*3*4-1*2*3+....+n(n+1)(n+2)-(n-1)n(n+1)]/3
[前后消项]
=[n(n+1)(n+2)]/3

所以1^2+2^2+3^2+......+n^2
=[n(n+1)(n+2)]/3-[n(n+1)]/2
=n(n+1)[(n+2)/3-1/2]
=n(n+1)[(2n+1)/6]
=n(n+1)(2n+1)/6