几题数学题,帮解一下thanks
3个回答
1.原式=2*0*(2*0+1)+2*1*(2*1+1)
.2*1005*(2*1005+1)
=2*1*(2*1+1).2*1005*(2*1005+1)
其通项公式=2*n(2*n+1)
=2*(2n^2+n)
=4n^2+2n
=4*(1^2+2^2.1005)^2+2(1+2+3.1005)
=4*[1005*1006*2011]/6+1005*1006
=1356465250
2.原式=(7-6)/6*7+(8-7)/7*8...+(101-100)/100*101
=(1/6)-(1/7)+(1/7)-(1/8).+(1/100)-(1/101)
=(1/6)-(1/101)
3.原式=(1+1/2)(1+1/3)(1+1/4)..(1+1/99)*(1-1/2)(1-1/3)...(1-1/99)
=(3/2)(4/3)(5/4)...(100/99)*(1/2)(2/3)*(3/4).(98/99)
=(100/2)*(1/99)
=50/99
谜语
1.公式
2.区间
