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Question

1+1+a2!+1+a+a23!+...=eaeab Find b

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Solution

In the L.H.S.first term contains one term, second term contains two terms, third term contains three terms,..., nth term contains n terms.
nth Term
Tn=1+a+a2+..+an1n!

=1.(1an)(1a)n!= 1(1a)(1n!ann!)
Hence sum of the series
S=n=1Tn
=n=11(1a)(1n!ann!)
=1(1a)(n=11n!n=1ann!)

=1(1a)((11!+12!+13!+...)(a1!+a22!+a33!+...))
=1(1a)((e1)(ea1))=eae(a1)

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