a1=2
a2=a1+2=4
a3=a2+5=9
a4=a3+8=17
a5=a4+11=28
a6=a5+14=42
2,5,8,11,14 ...是公差為3的等差數(shù)列
an=a(n-1)+2+(n-2)*3=a(n-1)+3n-4 (n>=2)
由上式可以求出an 從而可以求出Snan=a(n-1)+2+(n-2)*3=a(n-1)+3n-4 (n>=2)
由上式可以求出an
怎么求- -an=a(n-1)+3n-4
=a(n-2)+ (3n-4)+(3(n-1)-4)
=a(n-3)+(3n-4)+(3(n-1)-4)+(3(n-2)-4)
=....
=a(1)+(3n-4)+(3(n-1)-4)+(3(n-2)-4)+....+(3*2-4)
=2+3(n+(n-1)+(n-2)+.....+2) -4(n-1)
=2+3n(n+1)/2 -3-4n+4
=3n(n+1)/2 -4n+3
=(3n^2 +3n-8n)/2+3
=(3n^2-5n)/2+3