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

શ્રેણી \(\frac{{3 \times 1}}{{{1^2}}} + \frac{{5 \times ({1^3} + {2^3})}}{{{1^2} + {2^2}}} + \frac{{7 \times ({1^3} + {2^3} + {3^3})}}{{{1^2} + {2^2} + {3^2}}} + .......\) ના પ્રથમ \(10\) પદ સુધીનો સરવાળો મેળવો.

  1. A \(620\)
  2. B \(660\)
  3. C \(680\)
  4. D \(600\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(660\)

Step-by-step Solution

Detailed explanation

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