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(A) The binomial expansion of $(1+x)^{37}$ has $37+1 = 38$ terms. \ The coefficients are given by $\binom{37}{r}$ for $r = 0, 1, 2, \dots, 37$. \ Th

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$2^{36}$

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(A) The binomial expansion of $(1+x)^{37}$ has $37+1 = 38$ terms. \ The coefficients are given by $\binom{37}{r}$ for $r = 0, 1, 2, \dots, 37$. \ The total number of terms is $38$. The last $19$ terms correspond to $r = 19, 20, \dots, 37$. \ The sum of all coefficients is $\sum_{r=0}^{37} \binom{37}{r} = 2^{37}$. \ By symmetry,$\binom{37}{r} = \binom{37}{37-r}$. \ Let $S$ be the sum of the coefficients of the last $19$ terms: $S = \binom{37}{19} + \binom{37}{20} + \dots + \binom{37}{37}$. \ The sum of the first $19$ terms is $\binom{37}{0} + \binom{37}{1} + \dots + \binom{37}{18}$. \ Since $\binom{37}{0} = \binom{37}{37}, \binom{37}{1} = \binom{37}{36}, \dots, \binom{37}{18} = \binom{37}{19}$,the sum of the first $19$ terms is equal to the sum of the last $19$ terms. \ Thus,$2S = \sum_{r=0}^{37} \binom{37}{r} = 2^{37}$. \ Therefore,$S = \frac{2^{37}}{2} = 2^{36}$.

List-$I$	List-$II$
$(A)$ $a_0 + a_2 + \ldots + a_{2n}$	$(I)$ $n \cdot 3^{n-1}$
$(B)$ $a_1 + a_3 + \ldots + a_{2n-1}$	$(II)$ $n \cdot 3^n$
$(C)$ $a_1 + 2a_2 + 3a_3 + \ldots + 2n a_{2n}$	$(III)$ $\frac{1}{2}(3^n + 1)$
	$(IV)$ $\frac{1}{2}(3^n - 1)$

The sum of the coefficients of the last $19$ terms in the binomial expansion of $(1+x)^{37}$ is

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