If `I_(n)=(d^(n))/(dx^(n))(x^(n)lnx)`, then the value of `(1)/(50)(I_(7)-7I_(6))` is equal to

Answer

To solve the problem, we need to evaluate \( I_n = \frac{d^n}{dx^n} (x^n \ln x) \) and then find the value of \( \frac{1}{50}(I_7 - 7I_6) \).

### Step-by-Step Solution:

1. **Understanding \( I_n \)**:
\[ I_n = \frac{d^n}{dx^n} (x^n \ln x) \]
This represents the \( n \)-th derivative of the function \( x^n \ln x \).

2. **Using the Leibniz Rule**:
We can apply the Leibniz rule for differentiation of a product:
\[ \frac{d^n}{dx^n}(uv) = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}}{dx^{n-k}}(u) \frac{d^k}{dx^k}(v) \]
Here, let \( u = x^n \) and \( v = \ln x \).

3. **Calculating \( \frac{d^k}{dx^k}(\ln x) \)**:
The \( k \)-th derivative of \( \ln x \) can be computed using the formula:
\[ \frac{d^k}{dx^k}(\ln x) = \frac{(-1)^{k-1}(k-1)!}{x^k} \]
for \( k \geq 1 \), and \( \frac{d^0}{dx^0}(\ln x) = \ln x \).

4. **Calculating \( \frac{d^{n-k}}{dx^{n-k}}(x^n) \)**:
The \( (n-k) \)-th derivative of \( x^n \) is:
\[ \frac{d^{n-k}}{dx^{n-k}}(x^n) = \frac{n!}{(n-k)!} x^{n-k} \]

5. **Combining Results**:
Using the Leibniz rule:
\[ I_n = \sum_{k=0}^{n} \binom{n}{k} \frac{n!}{(n-k)!} x^{n-k} \frac{(-1)^{k-1}(k-1)!}{x^k} \]
Simplifying this gives:
\[ I_n = n! \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(n-k)!} \cdot \frac{1}{x} = n! \cdot \frac{1}{x} \sum_{j=0}^{n-1} \frac{(-1)^{j}}{j!} \]
where \( j = k-1 \).

6. **Finding \( I_7 \) and \( I_6 \)**:
From the pattern, we can derive:
\[ I_n = n! \left( \sum_{j=0}^{n-1} \frac{(-1)^j}{j!} \right) \]
Therefore:
\[ I_7 = 7! \cdot \left( \sum_{j=0}^{6} \frac{(-1)^j}{j!} \right) \]
\[ I_6 = 6! \cdot \left( \sum_{j=0}^{5} \frac{(-1)^j}{j!} \right) \]

7. **Calculating \( I_7 - 7I_6 \)**:
\[ I_7 - 7I_6 = 7! \left( \sum_{j=0}^{6} \frac{(-1)^j}{j!} \right) - 7 \cdot 6! \left( \sum_{j=0}^{5} \frac{(-1)^j}{j!} \right) \]
This simplifies to:
\[ I_7 - 7I_6 = 7! \cdot \frac{(-1)^6}{6!} = 7! \cdot \frac{1}{720} = 5040 \cdot \frac{1}{720} = 7 \]

8. **Final Calculation**:
\[ \frac{1}{50}(I_7 - 7I_6) = \frac{1}{50} \cdot 7 = \frac{7}{50} \]

### Final Answer:
\[ \frac{1}{50}(I_7 - 7I_6) = \frac{7}{50} \]