To find the sum of the last 30 coefficients in the binomial expansion of \((1 + x)^{59}\), we can use the formula for the binomial expansion:
\((1 + x)^{59} = \binom{59}{0} + \binom{59}{1}x + \binom{59}{2}x^2 + \ldots + \binom{59}{59}x^{59}\)
In this expansion, the coefficient of \(x^k\) term is given by \(\binom{59}{k}\).
We want to find the sum of the coefficients from \(x^{59}\) down to \(x^{30}\), inclusive.
So, the sum we need to find is:
\(\binom{59}{59} + \binom{59}{58} + \binom{59}{57} + \ldots + \binom{59}{30}\)
Now, recall that the sum of the coefficients in the expansion of \((1+x)^n\ is\ 2^n\).
So, in our case, the sum of all coefficients in the expansion of \((1+x)^{59}\ is \ 2^{59}\).
Therefore, the sum of the last 30 coefficients is half of the sum of all coefficients:
\(\frac{1}{2} \times 2^{59} = 2^{58}\)
So, the correct option is (C): \(2^{58}\)
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is