The sum of the coefficients of three consecutive terms in the binomial expansion of $$(1 + x)^{n+2}$$, which are in the ratio 1 : 3 : 5, is equal to

Three consecutive terms $$\binom{n+2}{r}$$, $$\binom{n+2}{r+1}$$, $$\binom{n+2}{r+2}$$ in the expansion of $$(1+x)^{n+2}$$ are in the ratio $$1:3:5$$.

From the first ratio:

$$\frac{\binom{n+2}{r+1}}{\binom{n+2}{r}} = \frac{n+2-r}{r+1} = 3$$

$$n + 2 - r = 3r + 3 \implies n = 4r + 1 \quad \cdots(1)$$

From the second ratio:

$$\frac{\binom{n+2}{r+2}}{\binom{n+2}{r+1}} = \frac{n+1-r}{r+2} = \frac{5}{3}$$

$$3(n+1-r) = 5(r+2) \implies 3n = 8r + 7 \quad \cdots(2)$$

Substituting (1) into (2):

$$3(4r+1) = 8r+7 \implies 12r+3 = 8r+7 \implies r = 1$$

$$n = 5$$, so the expansion is $$(1+x)^7$$.

The three consecutive terms:

$$\binom{7}{1} = 7, \quad \binom{7}{2} = 21, \quad \binom{7}{3} = 35$$

Ratio: $$7:21:35 = 1:3:5$$ ✓

Sum = $$7 + 21 + 35 = 63$$

The answer is Option B: $$63$$.