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The sum 1 + 1 + 32! + 1 + 3 + 53! + 1 + 3 + 5 + 74! + up to terms, is equal to
(2) 2e
Numerator of the n-th term = 1 + 3 + 5 + + (2n - 1) = n2
⇒ Tn = n2n! = n(n - 1)! = (n - 1) + 1(n - 1)!
⇒ Tn = 1(n - 2)! + 1(n - 1)! (first part only for n 2)
S = n=2 1(n - 2)! + n=1 1(n - 1)!
∵ r=0 1r! = e
∴ S = e + e = 2e
Answer: (2)
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