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JEE Mains · Maths · STD 11 - 8. sequence and series

निम्न श्रेणी \(1+6+\frac{9\left(1^{2}+2^{2}+3^{2}\right)}{7}+\frac{12\left(1^{2}+2^{2}+3^{2}+4^{2}\right)}{9}\) \(+\frac{15\left(1^{2}+2^{2}+\ldots .+5^{2}\right)}{11}+\ldots\) के प्रथम \(15\) पदों का योग है 

  1. A \(7820\)
  2. B \(7830\)
  3. C \(7520\)
  4. D \(7510\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(7820\)

Step-by-step Solution

Detailed explanation

\({T_n} = \frac{{\left( {3 + \left( {n - 1} \right) \times 3} \right)\left( {{1^2} + {2^2} + ... + {n^2}} \right)}}{{\left( {2n + 1} \right)}}\)…
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