Answer & Solution
Answer: Option B
Solution:
$$\eqalign{
& {S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2} \cr
& {\text{Here,}} \cr
& d = 1 \cr
& a = 1\, \cr
& {\text{and}}\,\,n - 1\,\,{\text{terms}} \cr} $$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + \left( {n - 1 - 1} \right)} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + n - 2} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)}}{2}$$
