Question

1*2+2*3+3*4+……+n（n+1）求和

高一数列问题

Answer 1

1、可以用公式求和
n(n+1)=n²+n
1*2+2*3+3*4+……+n（n+1）
=1+2²+3²+…+n²+1+2+3+…+n
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)(n+2)/3
2、可以用裂项求和
n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
1*2+2*3+3*4+……+n（n+1）
=[(1*2*3-0*1*2)+(2*3*4-1*2*3)+(3*4*5-2*3*4)+…+n(n+1)(n+2)-(n-1)n(n+1)]/3
=n(n+1)(n+2)/3

Answer 2

1*2+2*3+3*4+…+n(n+1)
=（1^2+1）+（2^2+2）+（3^2+3）+.......+（n^2+n）
=（1^2+2^2+3^2+.....+n^2）+（1+2+3+....+n）
=[n(n+1)(2n+1)/6]+[n(1+n)/2]
=n(n+1)(n+2)/3

Answer 3

n(n+1)=n^2+n

1*2+2*3+3*4+……+n（n+1）
=1^2+1+2^2+2+3^2+3+.....+n^2+n
=1^2+2^2+3^2+....+n^2+1+2+3+....+n
=n(n+1)(2n+1)/6+n(n+1)/2
=n(n+1)/2*[(2n+1)/3+1]
=n(n+1)/2*(2n+4)/3
=n(n+1)/2*2(n+2)/3
=n(n+1)(n+2)/3

Answer 4

an=n(n+1)=n^2+n
Sn=n(n+1)(2n+1)/6+n(n+1)/2=n(n+1)(2n+1+3)/6=n(n+1)(n+2)/3