(1) + (1 + 1) + (1 + 1 + 1) + ....... + (1 + 1 + 1 + ...... n - 1 times) = ......

Answer: Option B

$$\begin{array}{rl}& S_n=\left[2a+\left(n-1\right)d\right]\times \frac{n}{2}\\ & \text{Here,}\\ & d=1\\ & a=1\\ & \text{and}n-1\text{terms}\end{array}$$
$$\Rightarrow S_{n-1}=\left[2+\left(n-1-1\right)\right]$$     $$\times \frac{\left(n-1\right)}{2}$$
$$\Rightarrow S_{n-1}=\left[2+n-2\right]$$    $$\times \frac{\left(n-1\right)}{2}$$
$$=\frac{n\left(n-1\right)}{2}$$