题在图上,
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题在图上,
题在图上,
题在图上,
(1)
a1+2a2+3a3..+nan=n(n+1)(n+2)----(1)==>
a1+2a2+3a3..+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)----(2);
(2)-(1)==>(n+1)a(n+1)=3(n+1)(n+2)==>a(n+1)=3(n+2)==>an=3(n+1)==>an=3n+3;
Sn=3*1+3+3*2+3+...+3*n+3=3(1+2+3+..+n)+3n=3n(n+1)/2+3n=3n(n+3)/2;
(2)
an+1-an=2n;==>
an-an-1=2(n-1)
...
a2-a1=2(2-1)=2;
左右相加得:
an+1-a1=2(1+2+..+n);a1=0;==>
an+1=n(n+1)==>
an=n(n-1);