正项数列an中,a1=1,an+1-√an+1=an+√an.证明数列an为等差数列并求通项an
问题描述:
正项数列an中,a1=1,an+1-√an+1=an+√an.证明数列an为等差数列并求通项an
证明√an为等差数列
答
an+1-√an+1=an+√an
得an+1-an=√an+1+√an
即(√an+1+√an)(√an+1-√an)=√an+1+√an
则√an+1-√an=1
故{√an}是首项为√a1=1公差为1的等差数列
则√an=n
故an=n^2