n+1=bn^2-(n-2)bn + 3
=(bn-n)*bn+2bn+3
>=2n+3
>=n+1
由归纳假设可得bn>=n
若数列{bn}满足：bn+1=bn^2-(n-2)bn + 3,且b1≥1,n∈N*,用数学归纳法证明：bn≥n
1个回答
相关问题
-
设数列{bn}满足bn=n^2/2^(n+1),证明:bn
-
若数列bn满足b1=2,且bn+1=bn+2^n+n,求数列bn的通项公式.
-
已知数列{an},{bn}满足bn=an+1-an其中n=1,2,3,…．若bn+1bn-1=bn（n≥2）,且b1=1
-
已知数列﹛bn﹜满足b1﹦-1,bn﹢1﹦bn+（2n-1）（n∈N﹡）.求数列﹛bn﹜的通项公式bn.
-
数列bn满足b1=3/4,且3bn-bn-1=n,求bn
-
已知数列{bn}满足b1=-1,b(n+1)=bn+(2n-1),求bn
-
已知数列{an},{bn}满足bn=an+1-an,其中n=1,2,3,… 若bn+1*bn-1=bn,且b1=1,b2
-
高中题库{bn}满足b1=2,且4/bn+1bn-6/bn+1+3/bn=0(n>1或n=1)求数列｛bn}的通项公式
-
3.设数列{an}的前n项和Sn=2an-4(n∈N+),数列{bn}满足:bn+1=an+2bn,且b1=2.求{bn
-
已知数列{bn}满足bn=n^2/3^n,证明:bn≤4/9