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Find the sum of the series whose nth term is (2n - 1)2 ?
1. \(\frac{{n \cdot \left( {n - 1} \right) \cdot \left( {2n - 5} \right)}}{6}\)
2. \(\frac{{n \cdot \left( {n - 1} \right) \cdot \left( {2n - 7} \right)}}{6}\)
3. \(\frac{{n \cdot \left( {n + 1} \right) \cdot \left( {2n + 7} \right)}}{6}\)
4. \(\frac{{n \cdot \left( {2n + 1} \right) \cdot \left( {2n - 1} \right)}}{3}\)
5. None of these

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Correct Answer - Option 4 : \(\frac{{n \cdot \left( {2n + 1} \right) \cdot \left( {2n - 1} \right)}}{3}\)

Concept:

  • 1 + 2 + 3 + ….. + n \( = \;\sum n = \frac{{n\;\left( {n\; + 1} \right)}}{2}\)
  • 12 + 22 + 32 + ……. + n2 \( = \;\sum {n^2} = \frac{{n\;\left( {n\; + 1} \right)\;\left( {2n\; + \;1} \right)}}{6}\)
  • 13 + 23 + 33 + ….. + n3 \( = \;\sum n^3 = \frac{{n^2\;\left( {n\; + 1} \right)^2}}{4}\)

Calculation:

Given: an = (2n - 1)2

As we know that, \({S_n} = \sum {{a_n}} \)

\( \Rightarrow {S_n} = \;\mathop \sum \limits_{k = 1}^n {a_k} = \;\mathop \sum \limits_{k = 1}^n \left[ {4k^2 + 4k + 1} \right]\)

\( = \;4\mathop \sum \limits_{k = 1}^n k + \;4\mathop \sum \limits_{k = 1}^n {k^2} + \;\mathop \sum \limits_{k = 1}^n {1} \)

\( \Rightarrow {S_n} = \frac{{n \cdot \left( {2n + 1} \right) \cdot \left( {2n - 1} \right)}}{3}\)

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