Question

a1=1/2 a1+a2+a3+…an=n2an,求证:an=1/n(n+1)

Answer 1

a1+a2+a3+…a(n-1)+an=n^2an

n=n-1时:

a1+a2+a3+…a(n-1) =(n-1)^2*a(n-1)

两式相减

an=n^2*an-(n-1)^2a(n-1)

(n-1)(n+1)*an=(n-1)^2*a(n-1)

an =a(n-1)* n-1/n+1

an-1=a(n-2)* n-2/n

……

a1=1/2

上式相乘

an=1/n*(n+1)