Question

已知数列{an},其中a1=1.an+1=an+2n+5,求它的通项公式.

Answer 1

解：由题可得：
a1=1
a2=a1+2x1+5
a3=a2+2x2+5
……
an=a(n-1)+2x(n-1)+5
把迟陪以唯吵上所有式子相加并化简得：
an=2x[1+2+3+……+(n-1)]+5x(n-1)+1=n(n-1)+5n-4=n�指旦侍0�5+4n-4

Answer 2

a(n+1)=an+2n+5
a(n+1) -an = 2n+5
an -a(n-1) = 2n+3
an -a1 = 7+9+..+(2n+3)
= (n+5)(n-1)
an = n^2+4n-4