Let `A_n` be the area bounded by the curve `y = x^n(n>=1)` and the line `x=0, y = 0 and x =1/2`.If `sum_(n=1)^n (2^n A_n)/n=1/3` then find the value of n.

Answer

To solve the problem step by step, we will find the area \( A_n \) bounded by the curve \( y = x^n \), the lines \( x = 0 \), \( y = 0 \), and \( x = \frac{1}{2} \), and then use the given summation condition to find the value of \( n \).

### Step 1: Determine the area \( A_n \)

The area \( A_n \) can be calculated using the definite integral:

\[ A_n = \int_0^{1/2} x^n \, dx \]

### Step 2: Evaluate the integral

To evaluate the integral, we apply the power rule for integration:

\[ A_n = \left[ \frac{x^{n+1}}{n+1} \right]_0^{1/2} \]

Calculating this gives:

\[ A_n = \frac{(1/2)^{n+1}}{n+1} - 0 = \frac{1}{2^{n+1}(n+1)} \]

### Step 3: Substitute \( A_n \) into the summation

We are given the equation:

\[ \sum_{n=1}^{N} \frac{2^n A_n}{n} = \frac{1}{3} \]

Substituting \( A_n \):

\[ \sum_{n=1}^{N} \frac{2^n \cdot \frac{1}{2^{n+1}(n+1)}}{n} = \frac{1}{3} \]

This simplifies to:

\[ \sum_{n=1}^{N} \frac{1}{2(n+1)n} = \frac{1}{3} \]

### Step 4: Simplify the summation

We can rewrite \( \frac{1}{2(n+1)n} \) as:

\[ \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]

Thus, the summation becomes:

\[ \frac{1}{2} \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} \right) \]

This is a telescoping series, which simplifies to:

\[ \frac{1}{2} \left( 1 - \frac{1}{N+1} \right) \]

### Step 5: Set the equation equal to \( \frac{1}{3} \)

Now we set the expression equal to \( \frac{1}{3} \):

\[ \frac{1}{2} \left( 1 - \frac{1}{N+1} \right) = \frac{1}{3} \]

### Step 6: Solve for \( N \)

Multiplying both sides by 2:

\[ 1 - \frac{1}{N+1} = \frac{2}{3} \]

Rearranging gives:

\[ \frac{1}{N+1} = 1 - \frac{2}{3} = \frac{1}{3} \]

Taking the reciprocal:

\[ N + 1 = 3 \implies N = 2 \]

### Conclusion

Thus, the value of \( n \) is:

\[ \boxed{2} \]