jmc

Question

Let S be the sum of all the real coefficients of the expansion of (1+ix)^2009. What is _2(S)?

Accepted Answer

By the Binomial Theorem, (1 + ix)^2009 = 1 + 20091 ix - 20092 x^2 - 20093 ix^3 + 20094 x^4 + .Also, (1 - ix)^2009 = 1 - 20091 ix - 20092 x^2 + 20093 ix^3 + 20094 x^4 + .Adding the two, all the imaginary terms cancel, and we are left with the real terms: (1 + ix)^2009 + (1 - ix)^2009 = 2 [ 1 - 20092 x^2 + 20094 x^4 + ].Then we can find the sum of the real terms by dividing by 2 and setting x = 1: (1 + i)^2009 + (1 - i)^20092.We can write align* (1 + i)^2009 &= (1 + i) (1 + i)^2008 &= (1 + i) ((1 + i)^2)^1004 &= (1 + i) (1 + 2i - 1)^1004 &= 2^1004 (1 + i). align*Similarly, (1 - i)^2009 = 2^1004 (1 - i), so (1 + i)^2009 + (1 - i)^20092 = 2^1004 (1 + i) + 2^1004 (1 - i)2 = 2^1004.Therefore, _2 S = 1004.