If the complex number sequence {an} satisfies A1 = 0, an = [a (subscript) n-1] ^ 2 + I (n > = 2, I is an imaginary unit), then the sum of its first 2007 terms is Detailed process

If the complex number sequence {an} satisfies A1 = 0, an = [a (subscript) n-1] ^ 2 + I (n > = 2, I is an imaginary unit), then the sum of its first 2007 terms is Detailed process

a1=0
a2=a1^2+i=i
a3=i^2+i=i-1
a4=(i-1)^2+i=-i
a5=(-i)^2+i=i-1
a6=(i-1)^2+i=-i
.
a1+a2+a3+...+a2007=0+i+i-1-i+i+1-i.+i-1
=i-1002+i-1
=2i-1003

Let f (x) = A0 + a1x +... + anx ^ n be a polynomial with integral coefficients of degree n. if an, A0 and f (1) are all odd numbers, it is proved that f (x) = 0 has no rational root

Counter proof: suppose there is a rational root, let P / Q (P, q are coprime integers, and Q is not equal to 0), then (X-P / Q) | f (x), because f (x) is a polynomial with integral coefficients, and it is reducible in the rational number field, then we can get qx-p | f (x) [primitive polynomial science, if a non-zero polynomial with integral coefficients can be decomposed into the product of two polynomials with lower degree of rational coefficients, Then it must be decomposed into two integral coefficient polynomials of lower degree, where f (x) = (X-P / Q) g (x), it is deduced that f (x) = (qx-p) H (x) holds]. According to the theorem P | A0, Q | an, it can be seen that P and Q are both odd numbers, f (1) = (Q-P) H (1), and f (1) is odd number, H (1), which is an integer, then Q-P is odd number (odd number can only be the product of two odd numbers), and P and Q are odd numbers, Q-P must be even number

Let f (x) = A0 + a1x + a2x ^ 2 +... + anxn be a polynomial with integer coefficients of degree n. if an, A0 and f (1) are all odd numbers, it is proved that f (x) = 0 has no rational root

If R / S is the rational root of F (x), then f (x) = (sx-r) g (x), where g (x) is an integral coefficient polynomial. Because R | A0, s | an, and an, A0, are odd numbers, so r and s are odd numbers, so S-R is even number. So f (1) = (S-R) g (1) is even number, resulting in contradiction!