Correct Answer - Option 3 : 5n
2 + 7n / 2
Concept:
- 1 + 2 + 3 + ….. + n \( = \;\sum n = \frac{{n\;\left( {n\; + 1} \right)}}{2}\)
- 12 + 22 + 32 + ……. + n2 \( = \;\sum {n^2} = \frac{{n\;\left( {n\; + 1} \right)\;\left( {2n\; + \;1} \right)}}{6}\)
- 13 + 23 + 33 + ….. + n3 \( = \;\sum n^3 = \frac{{n^2\;\left( {n\; + 1} \right)^2}}{4}\)
Calculation:
Here, we have to find the sum of n terms of the infinite series 6 + 11 + 16 + ......
Here, the general term of the series is given by: an =
As we know that, \(S_n = \;\sum a_n\)
\( ⇒ {S_n} = \;\mathop \sum \limits_{k = 1}^n {a_k} = \;\mathop \sum \limits_{k = 1}^n \left[ {5k + 1} \right]\)
⇒ Sn = 5n2 + 7n / 2