If we know that in the expansion of (1 + ax) 6, the coefficient of X3 is equal to 160, then the real number a=______ ．

If we know that in the expansion of (1 + ax) 6, the coefficient of X3 is equal to 160, then the real number a=______ ．

The expansion of ∵ (1 + ax) 6 is tr + 1 = Cr6 · (AX) r, let r = 3, the coefficient with X3 term is A3 · C36 = 160, the solution is a = 2, so the answer is: 2

If the coefficient of X * 3 in the expansion of (1 + x) * 6 (1-ax) * 2 is 20, then the nonzero real number a=___

(1 + x) ^ 6 (1-ax) ^ 2 = (1 + x) ^ 6 (a ^ 2x ^ 2-2ax + 1) because the binomial expansion of (1 + x) ^ 6 is C (6, K) x ^ k, the coefficient of the X * 3 term in the expansion of (1 + x) * 6 (1-ax) * 2 is C (6,1) * a ^ 2 + C (6,2) * (- 2A) + C (6,3) * 1 = 6A ^ 2-30a + 20 = 20, because a is not equal to 0, so there is a = 5

On the function problem of binomial: an, BN is FN (x) = (1 + 2x) (1 + 2 ^ 2x) The coefficients of x ^ 2, X in (1 + 2 ^ NX) (n belongs to n *) expansion
(1) Find BN; (2) the recurrence formula of an; (3) an

(1) bn = 2 + 2^2 + ...+ 2^n
(2) an = bn^2 - 2^2 - 4^2 - 8^2 - ...- 2^2n
(3) It can be obtained from 1,2

Let the sum of the coefficients of the expansion of (3x ^ 1 / 3 + 2x ^ 1 / 2) ^ n be t, where the sum of the binomial coefficients is h, then t + H = 272, then the coefficients of the expansion of x ^ 2 are

Let x = 1, then t = 4 ^ n and H = 2 ^ n
4 ^ n + 2 ^ n = 272, n = 4
Then it is calculated by the general formula, that is, TR + 1 = C 3 ^ N-R 2 ^ R x ^ 1 / 3N + 1 / 6R
When n = 4, r = 4
The coefficient is C (4 4) 3 ^ 4-4 2 ^ 4 = 16

Judge whether the binary function of high number is continuous or not
x^2y/(x^2+y^2),x^2+y^2≠0
f(x,y)=
0 ,x^2+y^2=0
Is this function continuous at (0,0)? I'll repay you when I have money,

F (x, y) = x ^ 2Y / (x ^ 2 + y ^ 2), 0 ≤| f (x, y) | = x & # 178; / (X & # 178; + Y & # 178;) * | y ≤| y | LIM (x, y) - > (0,0) | y | = 0, Lim | f (x, y) | = 0, | Lim f (x, y) = 0 = f (0,0), so this function is continuous at (0,0). Note: calculate double limit

Judging the extremum of a function by its limit
Let Lim [f (x) - f (a)] / (x-a) ^ 2 be 1 when x tends to a, then f (x) is ()
(A) The derivative exists, but f '(a) is not equal to 1, and (b) has a maximum
(C) The minimum (d) derivative does not exist

First of all, Lim [f (x) - f (a)] / (x-a) ^ 2 = 1 when x tends to a, so there must be f (x) continuous at a point and lim [f (x) - f (a)] / (x-a) = 0, that is, f (x) differentiable at a point and f '(a) = 0. In fact, it is easy to prove that C is continuous at a point from F (x), Lim [f (x) - f (a)] / (x-a) ^ 2 is 1 when x tends to a