Correct Answer - Option 1 :
\(\frac{{n \cdot \left( {n + 1} \right) \cdot \left( {n + 2} \right)}}{3}\)
Concept:
-
1 + 2 + 3 + ….. + n \( = \;\sum n = \frac{{n\;\left( {n\; + 1} \right)}}{2}\)
- 12 + 22 + 32 + ……. + n2\( = \;\sum {n^2} = \frac{{n\;\left( {n\; + 1} \right)\;\left( {2n\; + \;1} \right)}}{6}\)
Calculation:
Here we have to find the sum if the n term of an infinite series: 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ......
Here , the nth term of the given series is given by: an = n ⋅ (n + 1)
So, the sum of the nth term of the given series is given by : \({S_n} = \;\mathop \sum \limits_{k = 1}^n {a_k} = \;\mathop \sum \limits_{k = 1}^n \left[ {k \cdot \left( {k + 1} \right)} \right]\)
\( = \;\mathop \sum \limits_{k = 1}^n k + \;\mathop \sum \limits_{k = 1}^n {k^2}\)
As we know that, \( \;\sum n = \frac{{n\;\left( {n\; + 1} \right)}}{2}\) and \( \;\sum {n^2} = \frac{{n\;\left( {n\; + 1} \right)\;\left( {2n\; + \;1} \right)}}{6}\)
\( \Rightarrow {S_n} = \frac{{n \cdot \left( {n + 1} \right) \cdot \left( {n + 2} \right)}}{3}\)