Examveda
Examveda

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

A. $$\frac{{n\left( {n + 1} \right)}}{2}$$

B. $$\frac{{n\left( {n - 1} \right)}}{2}$$

C. $${n^2}$$

D. $$n$$

Answer: Option B

Solution(By Examveda Team)

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

This Question Belongs to Arithmetic Ability >> Progressions

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