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

નીચે આપેલ શ્રેણીનો સરવાળો મેળવો. \(1 + 6 + \frac{{9({1^2} + {2^2} + {3^2})}}{7} + \frac{{12({1^2} + {2^2} + {3^2} + {4^2})}}{9} + \frac{{15({1^2} + {2^2} + .... + {5^2})}}{{11}} + ...\) \(15\) પદ સુધી 

  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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