I give a general conclusion: let the coefficient of the quadratic term of k+ 1 be the largest, then

(1) If n is odd, when k=(n- 1)/2 and k=(n+ 1)/2, the coefficients of the quadratic term are equal, and both are the largest terms;

② If n is an even number, k=n/2 is the largest term.

The proof is as follows: Let the maximum term be k+ 1 term, and use k≤(n+ 1)/2 to solve the inequality C(n, k)≥C(n, k- 1).

But because k is a natural number, the parity of n must be considered. If n is an odd number, k=(n+ 1)/2 is the maximum value, but the coefficients of k=(n- 1)/2 and k=(n+ 1)/2 are based on C (n, k).

If n is an even number, then k=(n+ 1)/2 is not an integer, so we have to take k=n/2. Complete the certificate.

Now back to the topic, we can see from the above conclusion that the maximum term satisfies n/2+ 1=4, so n=6.

The general term of (x 2+ 1/x 4) 6 expansion is c (6,k) * [(x 2) (6-k)] * [x (-4k)] = c (6,k) x (12. So the constant term is c (6 6,2) =15.

It is also possible that n equals 6. Next, we can use the general formula of binomial expansion. TR+ 1 = C6(r)x(2 *(6-r))* x(-4r)。

Or think of it as the multiplication of six formulas. First, take two x 2 from them, and there is a C6(2) method, that is, 15.