Question

1x3+3x5+5x7+7x9+ﾁ6￥8+97x99的算式

Answer 1

是求和的式子吗？
原式=(2^2-1)+(4^2-1)+……+(98^2-1)
=(2^2+4^2+ +98^2)-49
=4*(1^2+2^2+3^2+ ...+49^2)-49
=4*1/6*49*(49+1)(2*49-1)-49
如果是求通式的话
比如：97*99=(2*49-1)*(2*49+1)
即计算
1*3+3*5+5*7+...+(2*n-1)*(2*n+1)
=4*1/6*n*(n+1)(2*n-1)-n

Answer 2

an = (2n-1)(2n+1)
= 4n^2 -1
= 4n(n+1) - 4n -1
= (4/3)[ n(n+1)(n+2)-(n-1)n(n+1) ] - 2[n(n+1) -(n-1)n] - 1
Sn =a1+a2+...+an
= (4/3)n(n+1)(n+2)- 2n(n+1) - n

1x3+3x5+5x7+7x9+...+97x99
=S49
=(4/3)(49)(50)(51)- 2(49)(50) - 49
=200600-4900-49
= 195651