Question:
The series \( 1 + \frac{1 + x}{2!} + \frac{1 + x + x^2}{3!} + \frac{1 + x + x^2 + x^3}{4!} + \dots \) is equal to
The series \( 1 + \frac{1 + x}{2!} + \frac{1 + x + x^2}{3!} + \frac{1 + x + x^2 + x^3}{4!} + \dots \) is equal to
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The series \( 1 + \frac{x}{2!} + \frac{x^2}{3!} + \dots \) corresponds to the Taylor series expansion of \( e^x \) minus \( e \), and dividing by \( x - 1 \) gives the final result.
Updated On: Apr 22, 2026
- \( \frac{e^x + 1}{x - 1} \)
- \( \frac{e^x + 1}{x + 1} \)
- \( \frac{e^x - e}{x + 1} \)
- \( \frac{e^x - e}{x - 1} \)
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The Correct Option is D
Solution and Explanation
Step 1: Recognize the given series.
The given series is: \[ 1 + \frac{1 + x}{2!} + \frac{1 + x + x^2}{3!} + \frac{1 + x + x^2 + x^3}{4!} + \cdots \] This series is an extension of the Taylor series expansion of \( e^x \).
Step 2: Break the series into parts.
Notice that each term in the series has the form: \[ \frac{1 + x + x^2 + \dots + x^n}{n!} \] This can be interpreted as part of the expansion for \( e^x \) minus a constant, and the series is related to the expansion of \( e^x \) in the denominator.
Step 3: Identify the relationship with the Taylor series for \( e^x \).
The Taylor series for \( e^x \) is given by: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \] By subtracting \( e \) from both sides of this expansion: \[ e^x - e = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \] which resembles the series in the question.
Step 4: Use the series form for the result.
Now, we divide the result by \( x - 1 \) (since each term involves \( x \) and the denominator involves factorials), which transforms the series into: \[ \frac{e^x - e}{x - 1} \]
Step 5: Conclusion.
Thus, the series is equal to \( \frac{e^x - e}{x - 1} \), which corresponds to option (D).
Step 6: Verification.
The series can be verified by applying the expansion and comparing the results. The correct expression is indeed \( \frac{e^x - e}{x - 1} \), confirming that option (D) is correct.
The given series is: \[ 1 + \frac{1 + x}{2!} + \frac{1 + x + x^2}{3!} + \frac{1 + x + x^2 + x^3}{4!} + \cdots \] This series is an extension of the Taylor series expansion of \( e^x \).
Step 2: Break the series into parts.
Notice that each term in the series has the form: \[ \frac{1 + x + x^2 + \dots + x^n}{n!} \] This can be interpreted as part of the expansion for \( e^x \) minus a constant, and the series is related to the expansion of \( e^x \) in the denominator.
Step 3: Identify the relationship with the Taylor series for \( e^x \).
The Taylor series for \( e^x \) is given by: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \] By subtracting \( e \) from both sides of this expansion: \[ e^x - e = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \] which resembles the series in the question.
Step 4: Use the series form for the result.
Now, we divide the result by \( x - 1 \) (since each term involves \( x \) and the denominator involves factorials), which transforms the series into: \[ \frac{e^x - e}{x - 1} \]
Step 5: Conclusion.
Thus, the series is equal to \( \frac{e^x - e}{x - 1} \), which corresponds to option (D).
Step 6: Verification.
The series can be verified by applying the expansion and comparing the results. The correct expression is indeed \( \frac{e^x - e}{x - 1} \), confirming that option (D) is correct.
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