The sum of the series $$1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \ldots$$ upto 11th term is:

First we look at the terms one by one. We have $$1 ,\; 2 \times 3 ,\; 3 \times 5 ,\; 4 \times 7 ,\; \ldots$$ and so on. Observing the pattern, the first factor of the $$n^{\text{th}}$$ term is simply $$n$$ while the second factor is always one less than twice that $$n$$, namely $$(2n-1).$$

So the general $$n^{\text{th}}$$ term can be written as

$$T_n \;=\; n \,(2n-1).$$

Now we expand this product to make later addition easier:

$$T_n \;=\; n\,(2n-1) \;=\; 2n^2 - n.$$

We are asked to find the sum of the first eleven terms, so we must add $$T_1 + T_2 + \ldots + T_{11}.$$ Symbolically, that sum is

$$S_{11} \;=\; \sum_{n=1}^{11} \bigl(2n^2 - n\bigr).$$

Because summation distributes over addition (that is, $$\sum (A+B)=\sum A+\sum B$$), we separate the two parts:

$$S_{11} \;=\; 2 \sum_{n=1}^{11} n^2 \;-\; \sum_{n=1}^{11} n.$$

At this point we use two well-known formulae for consecutive integers.

First formula (sum of the first $$m$$ natural numbers):

$$\sum_{n=1}^{m} n \;=\; \frac{m(m+1)}{2}.$$

Second formula (sum of the squares of the first $$m$$ natural numbers):

$$\sum_{n=1}^{m} n^2 \;=\; \frac{m(m+1)(2m+1)}{6}.$$

Here, $$m = 11.$$ Substituting $$m=11$$ into the second formula gives

$$\sum_{n=1}^{11} n^2 \;=\; \frac{11 \times 12 \times 23}{6}.$$

We multiply step by step:

$$11 \times 12 = 132,$$

$$132 \times 23 = 3036,$$

$$\frac{3036}{6} = 506.$$

So we have

$$\sum_{n=1}^{11} n^2 = 506.$$

Next, substituting $$m=11$$ into the first formula gives

$$\sum_{n=1}^{11} n \;=\; \frac{11 \times 12}{2} \;=\; \frac{132}{2} \;=\; 66.$$

Now we return to our expression for $$S_{11}$$:

$$S_{11} \;=\; 2 \times 506 \;-\; 66.$$

First carry out the multiplication:

$$2 \times 506 = 1012.$$

Then perform the subtraction:

$$1012 - 66 = 946.$$

Thus the required sum of the first eleven terms is

$$S_{11} = 946.$$

Hence, the correct answer is Option C.