Step 1: Understanding the Concept:
This is a problem on reduction formulas in integral calculus.
Reduction formulas help us evaluate integrals involving an integer parameter \( n \) by relating \( I_n \) to \( I_{n-1} \).
The standard method to derive such relations is Integration by Parts (IBP).
Since we have an exponential function multiplied by a polynomial power, IBP is the ideal choice.
Step 2: Key Formula or Approach:
1. Integration by Parts: \( \int u \, dv = uv - \int v \, du \).
2. Let \( u = (x-1)^n \) (to be differentiated) and \( dv = e^x \, dx \) (to be integrated).
Step 3: Detailed Explanation:
Let's apply IBP to \( I_n = \int_0^1 e^x (x-1)^n dx \):
- \( u = (x-1)^n \implies du = n(x-1)^{n-1} dx \)
- \( dv = e^x dx \implies v = e^x \)
Applying the IBP formula:
\[ I_n = \left[ e^x (x-1)^n \right]_0^1 - \int_0^1 e^x \times n(x-1)^{n-1} dx \]
Notice that the remaining integral is exactly \( I_{n-1} \), with a coefficient \( n \):
\[ I_n = \left[ e^x (x-1)^n \right]_0^1 - n I_{n-1} \]
Now, evaluate the boundary term:
Upper limit (\( x=1 \)): \( e^1 (1-1)^n = e \times 0 = 0 \).
Lower limit (\( x=0 \)): \( e^0 (0-1)^n = 1 \times (-1)^n \).
Since the problem states \( n \) is an **odd** natural number, \( (-1)^n = -1 \).
So the boundary term calculation is: \( [0] - [1 \times (-1)] = 0 - (-1) = 1 \).
Substituting this back into our recurrence equation:
\[ I_n = 1 - n I_{n-1} \]
Rearranging to get the required expression:
\[ I_n + n I_{n-1} = 1 \]
This matches Option (A).
Step 4: Final Answer:
The sum \( I_n + n I_{n-1} \) is equal to \( 1 \).