Question 464918
If we were to expand the whole thing,


*[tex \LARGE (1+i)^{100} = (100C0) + (100C1)i - (100C2) - (100C3)i + (100C4) + ... - (100C98) - (100C99)i + (100C100)]


We see that the real part is


*[tex \LARGE Re((1+i)^{100}) = (100C0) - (100C2) + (100C4) - ... + (100C100)]


and the imaginary part is


*[tex \LARGE Im((1+i)^{100}) = (100C1)i - (100C3)i + (100C5)i - ... + (100C97)i - (100C99)i]


All of the imaginary terms cancel out, because we can group the terms together, noting that nCk = nC(n-k):


*[tex \LARGE ((100C1)i - (100C99)i) + (-(100C3)i + (100C97)i) + ... = 0 + 0 + 0 +... = 0]


Hence, we are left with the real part, which happens to be that real number you posted.