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Find the sum of n terms of the series: 6 + 11 + 16 + ...... ?
1. 5n2 + 7n / 3
2. 5n2 - 7n / 2
3. 5n2 + 7n / 2
4. 5n2 - 7n / 3
5. None of these

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Correct Answer - Option 3 : 5n2 + 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

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