Add brackets to the polynomial-a ^ 3 + 2A ^ 2-A + 1 as follows 1. Is the highest, the coefficient becomes positive 2. Make the coefficient of quadratic term positive 3. Put the odd items in the brackets with "-" at the front, and the rest items in the brackets with "+" at the front

Add brackets to the polynomial-a ^ 3 + 2A ^ 2-A + 1 as follows 1. Is the highest, the coefficient becomes positive 2. Make the coefficient of quadratic term positive 3. Put the odd items in the brackets with "-" at the front, and the rest items in the brackets with "+" at the front

1、
-(a³-2a²+a-1)
2、
The coefficient of the quadratic term is now a positive number
So it's - A & sup3; + 2A & sup2; - A + 1
3、
-(a³+a)+(2a²+1)

When a = - 1, the polynomial - A ^ 2n + 1 + A ^ 2n + 2A ^ 2n + 3 (n is a positive integer) is equal to
-a^(2n+1) + a^(2n) + 2a^(2n+3)

Is - A ^ (2n + 1) + A ^ (2n) + 2A ^ (2n + 3)
Or - A ^ (2n) + 1 + A ^ (2n) + 2A ^ (2n) + 3
Note: (- 1) ^ (2n) = [(- 1) ^ 2] ^ n = 1 ^ n = 1
If the former, then - A ^ (2n + 1) + A ^ (2n) + 2A ^ (2n + 3)
= -(-1)^(2n+1) + (-1)^(2n) + 2*(-1)^(2n+3)
= -(-1)^(2n) * (-1) + 1 + 2*(-1)^(2n) * (-1)^3
= -1 * (-1) + 1 + 2 * 1 * (-1) = 0
If the latter, then - A ^ (2n) + 1 + A ^ (2n) + 2A ^ (2n) + 3
=-(-1)^(2n) + 1 + (-1)^(2n) + 2(-1)^(2n) + 3
= -1 + 1 + 1 + 2 + 3 = 6

3kx + 6ky write out the common factor that can be extracted from the factorization factor of polynomial

3kx+6ky
=3k(x+2y)

∑(x^2n)*(-1)^(n-1)/n(2n-1)

Let f (x) = ∑ (1 - > ∞) (x ^ 2n) * (- 1) ^ (n-1) / N (2n-1) = ∑ (1 - > ∞) (x ^ 2n) * (- 1) ^ (n-1) / N (2n-1) f '(x) = ∑ (1 - > ∞) (2x ^ (2n-1)) * (- 1) ^ (n-1) / (2n-1) f' '(x) = ∑ (1 - > ∞) (2x ^ (2n-2)) * (- 1) ^ (n-1) = 2 ∑ (0 - > ∞) (x ^ (2n)) * (- 1) ^ n = 2 ∑ (0 - > ∞) ((x ^ 2