Question:
The sum
\[
1 + \frac{1+a}{2!} + \frac{1+a+a^2}{3!} + \cdots
\]
is equal to:
The sum
\[
1 + \frac{1+a}{2!} + \frac{1+a+a^2}{3!} + \cdots
\]
is equal to:
Show Hint
Break sums into known exponential series.
Updated On: Mar 23, 2026
- \(e^a\)
- \(\dfrac{e^a-e}{a-1}\)
- \((a-1)e^a\)
- (a+1)eᵃ
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write as difference of two exponential series and simplify to obtain (eᵃ-e)/(a-1).
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