Correct Answer - Option 3 :
\(\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}\)
Concept:
- 1 + 2 + 3 + .... + n = n(n + 1)/2
- 12 + 22 + 32 + .... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)
Calculation:
Here, we have to find the sum of the series 2 + 6 + 12 + 20 + 30 +.....+ upto n terms
The nth term of the given series Tn = n(n + 1) = n2 + n
⇒ \(\)\(2 + 6 + 12 + \ldots + {T_n} = \;\mathop \sum \limits_{k = 1}^n k\left( {k + 1} \right) = \;\mathop \sum \limits_{k = 1}^n {k^2} + \;\mathop \sum \limits_{k = 1}^n k\)
As we know that, 1 + 2 + 3 + .... + n = n(n + 1)/2 and 12 + 22 + 32 + .... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)
⇒ \(2 + 6 + 12 + \ldots + {T_n} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}\)
Hence, option 3 is the correct answer.