Question

Find the sum of the series 2 + 6 + 12 + 20 + 30 +.....+ upto n terms
1. n(n + 2)
2. \(\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{2}\)
3. \(\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}\)
4. n(n + 1)(n + 2)
5. None of these

Answer 1

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 T_n = 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.