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Algebra Difficulty 4.8 AIME Find the answer

6. Given that the real part of the expansion of (1+ix)n+2(xI)(1+i x)^{n+2}(x \in I) is a polynomial in xx, then the sum of the coefficients of this polynomial is (

Pick one

Solution

6. (B).

Let the real part of the expansion be f(x)f(x), and the imaginary part be g(x)g(x), then f(1)f(1) and g(1)g(1) are the sums of the coefficients of the real part polynomial and the imaginary part polynomial, respectively.
f(1)+ig(1)=(1+i)4n+2=22n+1(cosπ4+isinπ4)in+2=22n+1[cos(nπ+π2)+isin(nπ+π2)].=22n+1(1)ni \begin{array}{l} f(1)+i g(1)=(1+i)^{4 n+2} \\ =2^{2 n+1}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{i n+2} \\ =2^{2 n+1}\left[\cos \left(n \pi+\frac{\pi}{2}\right)+i \sin \left(n \pi+\frac{\pi}{2}\right)\right] . \\ =2^{2 n+1} \cdot(-1)^{n} i \end{array}

It is a pure imaginary number, so f(1)=0f(1)=0.

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