Maths Olympiad Prep

Track / Stage 3 / 169 of 260 #169 of 1964

Problem 169

AMC 10/12, early questions
Algebra Difficulty 3.4 Find the answer

The constant term in the expansion of (x2x)n(x- \frac{2}{x})^n is ______, given that only the fifth term has the maximum binomial coefficient in its expansion.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Since only the fifth term has the maximum binomial coefficient in the expansion of (x2x)n(x- \frac{2}{x})^n, it implies that nn is an even number.
The expansion has a total of 9 terms, which means n=8n=8.
For (x2x)n(x- \frac{2}{x})^n, i.e., (x2x)8(x- \frac{2}{x})^8, the general term formula of its expansion is Tr+1=C8rx8r(2x)r=C8r(2)rx82rT_{r+1} = C_{8}^{r}x^{8-r}(- \frac{2}{x})^{r} = C_{8}^{r}(-2)^{r}x^{8-2r},
Let 82r=08-2r=0, we find r=4r=4. Therefore, the constant term in the expansion is C84(2)4=1120C_{8}^{4}(-2)^4 = 1120.
Hence, the answer is 1120\boxed{1120}.
By determining n=8n=8 from the problem statement and setting the exponent of xx in the general term formula of the binomial expansion to zero, we can find the value of rr and thus calculate the value of the constant term in the expansion.
This problem mainly examines the application of the binomial theorem, the properties of binomial coefficients, the general term formula of binomial expansion, and how to find the coefficient of a certain term in the expansion, which is a basic question.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.