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A problem of remainder theorem When f (x) = 2x ^ 3 + ax ^ 2 + BX + 3 divided by (x-1) (x + 2), the remainder is (2x + 1) a) f(1),f(-2) b) Two equations with a and B as unknowns are established c) Find the values of a and B
Let f (x) = (2x + C) (x-1) (x + 2) + (2x + 1)
a,f（1）=3,f（-2）=-3
b,2+a+b+3=3,-16+4a-2b+3=-3
A + B = - 2, 2a-b = 5
c,a=1,b=-3
On the problem of theorem with quadratic term
What is the coefficient of the fifth power of X in the expansion of (x + 5x / 1) to the seventh power? Please tell me how to solve this problem! After I substitute it into the quadratic term theorem, there is a fractional form. I don't know how to simplify it
What is the coefficient of x ^ 5 in the expansion of (x + 1 / 5x) ^ 7?
R + 1 = T &; R + 1 &; = C (n, R) [x ^ (7-r)] [1 / 5x) ^ R = C (n, R) [(1 / 5) ^ R] [x ^ (7-2r)]
It is known that 7-2r = 5, so r = 1, that is, the second term contains x ^ 5, so its coefficient = C (7,1) × (1 / 5) = 7 / 5
A title of quadratic term theorem
It is known that (1 + x) + (1 + x) ^ 2 + (1 + x) ^ 3 +... + (1 + x) ^ n = A0 + A1 * x + A2 * x ^ 2 +... + an * x ^ n,
If a1 + A2 +... A (n-1) = 29-n
Then the positive integer n is equal to?
Why?
Given (1 + x) + (1 + x) ^ 2 + (1 + x) ^ 3 +... + (1 + x) ^ n = A0 + A1 * x + A2 * x ^ 2 +... + an * x ^ n, let x = 0, then 1 + 1 + +Let x = 1, then 2 + 2 ^ 2 +... + 2 ^ n = A0 + A1 +... + an = 2 * (2 ^ n-1) / (2-1) = 2 ^ (n + 1) - 2
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