How to find the maximum or minimum term of binomial coefficient expansion It's a coefficient, not a binomial coefficient,

According to the binomial theorem, write the general term, make the difference between two adjacent terms (equivalent to derivation), and find the extreme value

For example, the coefficients of a and B are 1

a. If the coefficient of B is 1, the maximum term of the coefficient should be one or two terms in the middle

In the expansion of (1 + 2x) ^ n, the coefficients of item 6 and item 7 are equal. Find the item with the largest binomial coefficient and the item with the largest coefficient in the expansion Ask for detailed steps !

The coefficients of item 6 and item 7 are equal, indicating that there are 12 items in total, n = 11, and the general formula is C (11, a) * (2x) ^ A. It is easy to know that the sixth and seventh items of binomial coefficient are the largest, that is, C (11,5) = C (11,6) = 462, and the eighth and ninth items have the largest coefficients
That is: C (11,7) * 2 ^ 7 = C (11,8) * 2 ^ 8 = 42240

How to find the binomial coefficient of constant term of (2x-1 / 2) ^ 6 expansion, such as this kind of problem-solving process, do you have any skills

The general formula of (2x-1 / 2) ^ 6 expansion is C (6, n) [(2x) ^ n] [(- 1 / 2) ^ (6-n)], and N is an integer of 0-6
Constant term is n = 0, C (6,0) [(2x) ^ 0] [(- 1 / 2) ^ (6-0)] = (- 1 / 2) ^ 6 = 1 / 64
The binomial formula is n = 2, C (6,2) [(2x) ^ 2] [(- 1 / 2) ^ (6-2)] = 15 * 4x ^ 2 * 1 / 16 = 14x ^ 2 / 4,
The binomial coefficient is 15 / 4
The general form of this kind of questions is (AX + b) ^ k, a and B are constants,
The general formula of expansion is C (k, n) [(AX) ^ n] [b ^ (K-N)], 0 ≤ n ≤ K, and N is an integer
The m-Term formula means that the degree of X is m, that is, n = m, and the degree of constant term is k-m
If the binomial, then n = 2, the degree of constant term is K-2

Removable discontinuities for limit problems of higher numbers
Is the discontinuity of y = sinxsin (1 / x) 0? Is it accessible? How to consider 1 / x? Isn't 1 / 0 meaningless?

Because the limit of Y is 0 when x tends to 0

The second kind of discontinuity is that at least one of the left and right limits of a function does not exist, so does the limit of infinity exist?

When the calculation result of limit is infinite, we generally don't say that the limit of function is infinite, but when x approaches to what point, the value of function approaches to infinity. It is a process, not a value, so it is not a limit. We can't say that the limit is infinite

Let f (x) = x ^ a sin1 / x, if x ≠ 0; = 0, if x = 0
Under what conditions can a make f (x) at point x = 0
1) Continuous; 2) derivable

lim x->0 x^a sin(1/x)=f(0)
Sin (infinity) belongs to [0,1]
So limx - > 0 x ^ a = 0
a>0
F (x) continuous
2）
limx->0(x^a sin(1/x))’=f'(0)
x^a sin(1/x)'=ax^(a-1)sin(1/x)-cos(1/x)*(x)^(-2)x^a
=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
f'(0)=lim h->0 [f(h)-f(0)]/h
=limh->0 h^a sin(1/h)/h
=limh->0 h^(a-1) sin(1/h)
So we need to be satisfied
lim x->0 ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)=limh->0 h^(a-1)sin(1/h)
If a

The reason why the high number x = 0 is the breaking point of F (x) = [x] sin1 / X

In the second class, there is no definition of the function at this point. When the independent variable approaches this point, the value of the function changes infinitely many times between two constants. For example, the function y = sin (1 / x) is at x = 0. This is the same

Find Lim X - > 0 ((e ^ x + e ^ 2x +... + e ^ NX) / N) ^ (E / x)
The answer is e ^ ((n + 1) / 2 * e), but I'm not right about the result. My way to get e ^ e is to use the sum formula of the equal ratio sequence at first, then make the number equivalent, and then use the second of the two important limits to calculate. Why not?

Use the second formula of the important limit directly. Then use the norbid rule directly. No sum is required

Find LIM ((e ^ x + e ^ 2x + e ^ 3x E ^ NX) / N) ^ (1 / x), n is a given natural number, and the constraint condition under Lim is x ~ 0

We can use the equivalent infinitesimal ln (1 + x) = x and the law of lobita,
Its limit is e ^ (n + 1) / 2
Original formula = exp {Lim {1 / X * ln [1 + (e ^ x + e ^ 2x +... + e ^ nx-n) / N]}}
x->0
=Exp [LIM (e ^ x + e ^ 2x +... + e ^ nx-n) / NX] - type 0 / 0
x->0
=exp[lim(e^x+2e^2x+...+ne^nx)/n]
x->0
=Exp (n + 1 / 2) --- e ^ x = 1 when X - > 0
That is, its limit is e ^ [(n + 1) / 2]
This is 1991's mathematics three's entrance examination graduate school original question

How to find the maximum or minimum term of binomial coefficient expansion It's a coefficient, not a binomial coefficient,