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Mathematics Question on Sum of First n Terms of an AP

The sum 1+1+a2!+1+a+a23!+...∞1+\frac{1+a}{2!} + \frac{1+a+a^{2}}{3!} + ... \infty is equal to

A

eae^a

B

ea−ea−1\frac{e^{a} -e}{a-1}

C

(a−1)ea(a- 1) e^a

D

(a+1)ea( a +1)e^a

Answer

ea−ea−1\frac{e^{a} -e}{a-1}

Explanation

Solution

The given series is 1+1+a2!+1+a+a23!+1+a+a2+a34!+.....1+ \frac{1+a}{2!} + \frac{1+a+a^{2}}{3!} + \frac{1+a+a^{2} +a^{3}}{4!} + ..... Here , Tn=1+a+a2+a3+...to n termsn!T_{n} = \frac{1+a+a^{2}+a^{3} +...\text{to} \, n \, \text{terms}}{n!} =1(1−an)(1−a)(n!)=11−a(1−ann!) = \frac{1\left(1-a^{n}\right)}{\left(1-a\right)\left(n!\right)} = \frac{1}{1-a} \left(\frac{1-a^{n}}{n!}\right) ∴T1+T2+T3+....to ∞ \therefore T_{1} + T_{2} + T_{3} + .... \text{to } \infty =11−a[1−a1!+1−a22!+1−a33!+...to ∞]= \frac{1}{1-a} \left[\frac{1-a}{1!} + \frac{1-a^{2}}{2!} + \frac{1-a^{3}}{3!} + ... \text{to } \infty\right] =11−a[(11!+12!+13!+...to ∞)−(a1!+a22!+a33!+....to ∞)] = \frac{1}{1-a} \left[\left(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... \text{to} \, \infty\right) - \left(\frac{a}{1!} + \frac{a^{2} }{2!} + \frac{a^{3}}{3!} + ....\text{to} \, \infty\right)\right] =11−a[(e−1)−(ea−1)]= \frac{1}{1-a}\left[\left(e-1\right) - \left(e^{a}-1\right)\right] =e−ea1−a=ea−ea−1=\frac{e-e^{a}}{1-a} = \frac{e^{a}-e}{a-1}