求：1+3+6+10+15+21……=Sn?（用n表示）并证明

解：由数字规律得通项为1+2+3+…+n= n(n+1)/2

1

1+2

1+2+3

1+2+3+4

………

1+2+3+4+5… (n-1)

1+2+3+4+5…+ (n-1) +n

竖向相加Sn=1(n)+2(n-1)+3(n-2)+4(n-3)…+ (n-1)[n-(n-2)]+n[n-(n-1)]

Sn=1(n)+2(n-1)+3(n-2)+4(n-3)…+ (n-1)[n-(n-2)]+n[n-(n-1)]

Sn=n+2n+3n+4n+(n-1)n+n²+[-2-2*3-4*3-......-(n-1) (n-2)- n (n-1)]

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

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

Sn=n(1+2+3....n)-2[Sn- n (n+1)/2]

Sn=n²(n+1)/2-2[Sn- n (n+1)/2]

3Sn= n²(n+1)/2 +n (n+1)=n(n+1)(n+2)/2

Sn=n(n+1)(n+2)/2*3