The sum of the series 13!+25!+37!+... to ∞ is equal to
A
12e
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B
12e
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C
32e
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D
none of these
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Solution
The correct option is B12e Tn=n(2n+1)! =12[2n+1−1(2n+1)!] =12(2n)!−12(2n+1)! Applying summation, we get =12[(12!+14!+16!+...∞)−(13!+15!+...∞)] =12[12!−13!+14!−15!+...∞] =12.e−1 =12e.