JEE Math Practice Question
The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :
- A.(11)!
- B.10 $$ \times $$ (11!)
- C.101 $$ \times $$ (10!)
- D.11 $$ \times $$ (11!)
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