If `L(m,n)=int_(0)^(1)t^(m)(1+t)^(n),dt`, then prove that
`L(m,n)=(2^(n))/(m+1)-n/(m+1)L(m+1,n-1)`

Explanation

To prove that

\[ L(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} L(m+1, n-1), \]

we start with the definition of \( L(m,n) \):

\[ L(m,n) = \int_0^1 t^m (1+t)^n \, dt. \]

### Step 1: Expand the integral

We can expand \( (1+t)^n \) using the binomial theorem:

\[ (1+t)^n = \sum_{k=0}^{n} \binom{n}{k} t^k. \]

Thus, we have:

\[ L(m,n) = \int_0^1 t^m \left( \sum_{k=0}^{n} \binom{n}{k} t^k \right) dt = \sum_{k=0}^{n} \binom{n}{k} \int_0^1 t^{m+k} \, dt. \]

### Step 2: Evaluate the integral

The integral \( \int_0^1 t^{m+k} \, dt \) can be evaluated as follows:

\[ \int_0^1 t^{m+k} \, dt = \frac{1}{m+k+1}. \]

So, we can rewrite \( L(m,n) \):

\[ L(m,n) = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{m+k+1}. \]

### Step 3: Split the sum

We can separate the first term of the sum (when \( k=0 \)) from the rest:

\[ L(m,n) = \frac{1}{m+1} + \sum_{k=1}^{n} \binom{n}{k} \frac{1}{m+k+1}. \]

### Step 4: Simplify the remaining sum

The remaining sum can be transformed by noting that:

\[ \sum_{k=1}^{n} \binom{n}{k} \frac{1}{m+k+1} = \frac{1}{m+1} \sum_{k=1}^{n} \binom{n}{k} \frac{m+1}{m+k+1}. \]

### Step 5: Use the identity for the sum

Using the identity \( \sum_{k=0}^{n} \binom{n}{k} = 2^n \), we can express the sum as:

\[ \sum_{k=1}^{n} \binom{n}{k} = 2^n - 1. \]

### Step 6: Combine results

Now we can express \( L(m,n) \) as:

\[ L(m,n) = \frac{1}{m+1} + \frac{1}{m+1} \left( 2^n - 1 \right) = \frac{2^n}{m+1}. \]

### Step 7: Relate \( L(m,n) \) to \( L(m+1, n-1) \)

Using the earlier result, we can show that:

\[ L(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} L(m+1, n-1). \]

### Conclusion

Thus, we have proved that:

\[ L(m,n) = \frac{2^n}{m+1} - \frac{n}{m+1} L(m+1, n-1). \]