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Question

The sum of the series n=1n2+6n+10(2n+1)! is equal to :

A
418e+198e110
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B
418e+198e110
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C
418e198e110
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D
418e+198e1+10
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Solution

The correct option is C 418e198e110
n=1n2+6n+10(2n+1)!
Put 2n+1=r, where r=3,5,7....
n=r12n2+6n+10(2n+1)!=(r12)2+3r3+10r!=r2+10r+294(r!)
Now
r=3,5,7...r(r1)+11r+294(r!)=14r=3,5,7...(1(r2)!+11(r1)!+29r!)=14{(11!+13!+15!....)+11(12!+14!+16!.....)+29(13!+15!+17!+....)}=14{e1e2+11(e+1e22)+29(e1e22)}=18{e1e+11e+11e22+29e29e58}=18{41e19e80}
418e198e110

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