数列{an}是等差数列满足bn=a²n+1-an²,(1)求证bn是等差数列,(2)若{an}的d为8,b1=16,求Sn

数列{an}是等差数列满足bn=a²n+1-an²,(1)求证bn是等差数列,(2)若{an}的d为8,b1=16,求Sn
数列{an}是等差数列满足bn=a²n+1-an²,(1)求证bn是等差数列,(2)若{an}的d为8,b1=16,求Sn

数列{an}是等差数列满足bn=a²n+1-an²,(1)求证bn是等差数列,(2)若{an}的d为8,b1=16,求Sn
a(n)=a+(n-1)d,
b(n)=[a(n+1)]^2 - [a(n)]^2 = [a+nd]^2 - [a+(n-1)d]^2 = [a+nd+a+(n-1)d][a+nd-a-(n-1)d]
=d[2a+(2n-1)d]
=2ad+(2n-1)d^2
=2ad+d^2 + (2n-2)d^2
=(2ad+d^2) + (n-1)(2d^2),
{b(n)}是首项为(2ad+d^2),公差为(2d^2)的等差数列.
d=8时,
b(n)=(2a*8+64) + (n-1)(2*64) = 16(a+4) + 128(n-1),
16=b(1)=16(a+4),a=-3.
b(n)=16+128(n-1),
b(1)+b(2)+...+b(n)=16n + 128n(n-1)/2 = 16n + 64n(n-1).
a(n)=-3+8(n-1),
a(1)+a(2)+...+a(n)=-3n+8n(n-1)/2=-3n+4n(n-1).