Key Concepts and Formulas

• Integration by Parts: The formula for integration by parts is $\int u \, dv = uv - \int v \, du$. This is a crucial technique for finding recurrence relations for integrals.
• Definite Integral Properties: We will use the properties of definite integrals, particularly the evaluation at the limits of integration.
• Recurrence Relations: The problem is designed to establish a relationship between $I_n$ and $I_{n+1}$ (or $I_{n-1}$) to solve for the unknown parameter $k$.

Step-by-Step Solution

Step 1: Define the integral and the given relation. We are given the integral $I_n = \int_0^1 (1-x^k)^n dx$, where $k, n \in \mathbb{N}$. We are also given the relation $147 I_{20} = 148 I_{21}$.

Step 2: Establish a recurrence relation for $I_n$. We will use integration by parts on $I_n$. Let's rewrite $I_n$ as: $I_n = \int_0^1 (1-x^k)^n \cdot 1 \, dx$

To apply integration by parts, we need to choose $u$ and $dv$. A common strategy for integrals of the form $\int (f(x))^n g(x) dx$ is to let $u = (f(x))^n$. Let $u = (1-x^k)^n$ and $dv = dx$. Then, $du = n(1-x^k)^{n-1} \cdot (-kx^{k-1}) \, dx = -nk x^{k-1} (1-x^k)^{n-1} \, dx$. And $v = x$.

Applying the integration by parts formula: $I_n = \left[ x (1-x^k)^n \right]_0^1 - \int_0^1 x \left( -nk x^{k-1} (1-x^k)^{n-1} \right) \, dx$

Evaluate the first term: $\left[ x (1-x^k)^n \right]_0^1 = (1 \cdot (1-1^k)^n) - (0 \cdot (1-0^k)^n) = (1 \cdot 0^n) - (0 \cdot 1^n) = 0 - 0 = 0$.

Now, simplify the integral term: $I_n = 0 - \int_0^1 -nk x^k (1-x^k)^{n-1} \, dx$ $I_n = nk \int_0^1 x^k (1-x^k)^{n-1} \, dx$

Step 3: Manipulate the integral to relate it to $I_{n-1}$ or $I_n$. The integral term is $\int_0^1 x^k (1-x^k)^{n-1} \, dx$. We want to express this in terms of $I_{n-1}$ or $I_n$. Notice that $(1-x^k)^{n-1}$ is part of the expression for $I_{n-1}$. We can rewrite $x^k$ in terms of $(1-x^k)$ to get closer to the form of $I_{n-1}$. Let's add and subtract 1 inside the integral: $x^k = -(1-x^k) + 1$

Substitute this into the integral: $\int_0^1 x^k (1-x^k)^{n-1} \, dx = \int_0^1 [-(1-x^k) + 1] (1-x^k)^{n-1} \, dx$ $= \int_0^1 -(1-x^k)^n \, dx + \int_0^1 (1-x^k)^{n-1} \, dx$ $= - \int_0^1 (1-x^k)^n \, dx + \int_0^1 (1-x^k)^{n-1} \, dx$ $= -I_n + I_{n-1}$

Step 4: Substitute the result back into the recurrence relation for $I_n$. We found $I_n = nk \int_0^1 x^k (1-x^k)^{n-1} \, dx$. Substitute the expression from Step 3: $I_n = nk (-I_n + I_{n-1})$ $I_n = -nk I_n + nk I_{n-1}$

Rearrange to solve for $I_n$ in terms of $I_{n-1}$: $I_n + nk I_n = nk I_{n-1}$ $I_n (1 + nk) = nk I_{n-1}$ $I_n = \frac{nk}{1+nk} I_{n-1}$

This is a recurrence relation. We can also write it as: $\frac{I_n}{I_{n-1}} = \frac{nk}{1+nk}$

Step 5: Use the given relation to find $k$. The given relation is $147 I_{20} = 148 I_{21}$. This can be rewritten as $\frac{I_{21}}{I_{20}} = \frac{147}{148}$.

Now, let's use our recurrence relation for $n=21$. We have: $\frac{I_{21}}{I_{20}} = \frac{21k}{1+21k}$

Equate the two expressions for $\frac{I_{21}}{I_{20}}$: $\frac{147}{148} = \frac{21k}{1+21k}$

Step 6: Solve for $k$. Cross-multiply: $147 (1+21k) = 148 (21k)$ $147 + 147 \times 21k = 148 \times 21k$

Let $X = 21k$. The equation becomes: $147 + 147X = 148X$ $147 = 148X - 147X$ $147 = X$

Substitute back $X = 21k$: $147 = 21k$ $k = \frac{147}{21}$

To simplify $\frac{147}{21}$: $147 = 7 \times 21$ So, $k = \frac{7 \times 21}{21} = 7$.

Step 7: Verify the value of $k$. We found $k=7$. The problem states that $k \in \mathbb{N}$, and $7$ is a natural number.

Let's check if $k=7$ satisfies the given condition $147 I_{20} = 148 I_{21}$. The recurrence relation is $I_n = \frac{nk}{1+nk} I_{n-1}$. For $n=21$ and $k=7$: $I_{21} = \frac{21 \times 7}{1 + 21 \times 7} I_{20}$ $I_{21} = \frac{147}{1 + 147} I_{20}$ $I_{21} = \frac{147}{148} I_{20}$

Rearranging this gives: $148 I_{21} = 147 I_{20}$ This matches the given condition.

Common Mistakes & Tips

• Incorrectly choosing $u$ and $dv$ for integration by parts: The choice of $u$ and $dv$ is critical for obtaining a useful recurrence relation. Often, setting $u$ to the term with the higher power of $n$ helps.
• Algebraic errors when manipulating the integral: Be careful when rewriting terms like $x^k$ in terms of $(1-x^k)$ and when simplifying the resulting integrals.
• Mistakes in solving the final equation for $k$: Ensure careful cross-multiplication and algebraic manipulation when solving the equation derived from the given relation and the recurrence.

Summary

The problem requires finding a parameter $k$ by establishing a recurrence relation for the given definite integral $I_n$. We used integration by parts to relate $I_n$ to $I_{n-1}$. Specifically, by setting $u = (1-x^k)^n$ and $dv = dx$, we derived the recurrence $I_n = \frac{nk}{1+nk} I_{n-1}$. Using the given condition $147 I_{20} = 148 I_{21}$, which implies $\frac{I_{21}}{I_{20}} = \frac{147}{148}$, we substituted $n=21$ into the recurrence relation to get $\frac{I_{21}}{I_{20}} = \frac{21k}{1+21k}$. Equating the two expressions for the ratio $\frac{I_{21}}{I_{20}}$ and solving for $k$ yielded $k=7$.

The final answer is \boxed{7}.