Question

S=1,1+2,1+2+3,1+2+3+4,1+2+3+4.....+n 求Sn的前n项和的解题过程。。。要完整的。

Answer 1

1+2+3+4.....+n=n(n+1)/2=1/2(n^2+n).

1=1/2(1^2+1).
1+2=1/2(2^2+2).
1+2+3=1/2(3^2+3).
1+2+3+4=1/2(4^2+4).
…………

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

Answer 2

Sn-Sn-1=n(n+1)/2=0.5(n^2+n)
S2-S1=0.5(2^2+2)
S3-S2=0.5(3^2+3)
……
Sn-Sn-1=0.5(n^2+n)
累加得：Sn-S1=0.5[(1+2^2+3^2+……+n^2)+(1+2+3+4.....+n )]
Sn=S1+0.5[n(n+1)(2n+1)/6+n(n+1)/2]=n(n+1)(n+2)/3+1

Answer 3

Sn-Sn-1=n（n+1）/2然后依次加起来可求