The sum of the series $$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \ldots + 62 \cdot {}^{20}C_{20}$$ is equal to:

We observe that every term of the given series has the general form $$\bigl(2+3k\bigr)\,{}^{20}C_k$$ where the index $$k$$ runs from $$0$$ to $$20$$. Indeed, for $$k=0$$ we get $$2\,{}^{20}C_0$$, for $$k=1$$ we get $$\bigl(2+3\bigr)\,{}^{20}C_1=5\,{}^{20}C_1$$, and so on up to $$k=20$$ which gives $$\bigl(2+3\cdot20\bigr)\,{}^{20}C_{20}=62\,{}^{20}C_{20}$$.

So we can write the required sum as

$$S=\sum_{k=0}^{20}\bigl(2+3k\bigr)\,{}^{20}C_k.$$

Now we split this into two separate sums:

$$S=2\sum_{k=0}^{20}{}^{20}C_k+3\sum_{k=0}^{20}k\,{}^{20}C_k.$$

First we evaluate $$\displaystyle\sum_{k=0}^{20}{}^{20}C_k$$. By the binomial theorem we know the identity

$$(1+1)^{20}=\sum_{k=0}^{20}{}^{20}C_k\,1^{20-k}\,1^{k}$$

which simplifies to

$$\sum_{k=0}^{20}{}^{20}C_k=2^{20}.$$

Next we need $$\displaystyle\sum_{k=0}^{20}k\,{}^{20}C_k$$. A standard combinatorial identity gives

$$\sum_{k=0}^{n}k\,{}^{n}C_k=n\,2^{\,n-1},$$

so for $$n=20$$ we obtain

$$\sum_{k=0}^{20}k\,{}^{20}C_k=20\cdot2^{19}.$$

Substituting these two results back into the expression for $$S$$, we get

$$S=2\cdot2^{20}+3\cdot\bigl(20\cdot2^{19}\bigr).$$

Simplifying the first product, we have $$2\cdot2^{20}=2^{21}.$$ For the second product, we first multiply the constants: $$3\times20=60,$$ giving $$60\cdot2^{19}.$$ Hence

$$S=2^{21}+60\,2^{19}.$$ Now we factor out the common power $$2^{19}$$:

$$S=2^{19}\bigl(2^{2}+60\bigr)=2^{19}\bigl(4+60\bigr)=2^{19}\times64.$$

Recognising that $$64=2^{6}$$, we conclude

$$S=2^{19}\times2^{6}=2^{25}.$$

Hence, the correct answer is Option B.