Let a ≠ 0 and f (x) = a (x ^ + 1) - (2x + 1 / a) have the minimum value - 1, let the first n terms of an and Sn = f (n), let BN = (A2 + A4 +...) +a2n)/n,

Let a ≠ 0 and f (x) = a (x ^ + 1) - (2x + 1 / a) have the minimum value - 1, let the first n terms of an and Sn = f (n), let BN = (A2 + A4 +...) +a2n)/n,

X

Let the real number a ≠ 0, and the function f (x) = a (X & sup2; + 1) - (2x + 1 / a) have the minimum value - 1, let the first n terms of the sequence {an} and Sn = f (n), let BN = A2 + A4
Let the first n terms of the sequence {an} and Sn = f (n), let BN = (A2 + A4 +... + A2N) / N, n = 1,2,3..., prove that {BN} is an arithmetic sequence

Function and sequence synthesis according to the quadratic function has the minimum value - 1, we can get [4a (A-1 / a) - 4] / 4A = - 1, the solution is a = 1, a = - 2 (rounding off) because the minimum value, opening up, a > 0 to find an, using the formula, an = sn-sn-1, so bring in an = 2An + 2-A, so an = A-2 = - 1 to prove that BN is equal difference, through BN + 1-bn = constant BN = (A2

If the first term of the arithmetic sequence {an} is 1 and the sum of the first 10 terms is 145, then A2 + A4 + A8 + +a(2^n)=

S10 = 10 * a1 + 10 * (10-1) * D / 2
145=10*1+10*9*d/2,∴d=3
an=3n-2
a2+a4+a8+…… +a(2^n)
=3*2-2+3*4-2+3*8-2+…… +3*2^n-2
=3*（2+4+8+…… +2^n）-2n 2,4,8,…… , 2 ^ n equal ratio
=3*[2(1-2^n)/(1-2) ]-2n
=6*2^n-2n-6