The correct option is C 418e−198e−1−10
∞∑n=1n2+6n+10(2n+1)!
Put 2n+1=r, where r=3,5,7....
⇒n=r−12∴n2+6n+10(2n+1)!=(r−12)2+3r−3+10r!=r2+10r+294(r!)
Now
∑r=3,5,7...r(r−1)+11r+294(r!)=14∑r=3,5,7...(1(r−2)!+11(r−1)!+29r!)=14{(11!+13!+15!....)+11(12!+14!+16!.....)+29(13!+15!+17!+....)}=14{e−1e2+11(e+1e−22)+29(e−1e−22)}=18{e−1e+11e+11e−22+29e−29e−58}=18{41e−19e−80}
418e−198e−1−10