Question

数列{ an}中，a1=6,且an-an-1=an-1/n+n+1(n属于正整数，n大于等于2），则这个数列的通项公式an等于多少

Answer 1

解:
由于:
an-a(n-1)=a(n-1)/n+(n+1)
则:an=[(n+1)/n]*a(n-1)+(n+1)
则:an/(n+1)=a(n-1)/n
+1
设bn=an/(n+1)
则:b(n)
-b(n-1)=1
则{bn}为公差为1的等差数列
则:bn=b1+(n-1)*1=a1/2
+(n-1)=n+2
由于:bn=an/(n+1)=(n+2)
则:an=(n+1)(n+2)=n^2+3n+2