The value of `(.^(n)C_(0))/(n)+(.^(n)C_(1))/(n+1)+(.^(n)C_(2))/(n+2)+"..."+(.^(n)C_(n))/(2n)`
(a)`underset(0)overset(1)intx^(n-1)(1+x)^(n)dx` (b)`underset(1)overset(2)intx^(n)(x-1)^(n-1)dx` (c)`underset(1)overset(2)int(1+x)^(n)dx`
(d)`underset(0)overset(1)int(1-x)^(n)x^(n-1)dx`

Options

A. `underset(0)overset(1)intx^(n-1)(1-x)^(n)dx`
B. `underset(1)overset(2)intx^(n)(x-1)^(n-1)dx`
C. `underset(1)overset(2)int(1+x)^(n)dx`
D. `underset(0)overset(1)int(1-x)^(n)x^(n-1)dx`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \frac{{nC_0}}{n} + \frac{{nC_1}}{n+1} + \frac{{nC_2}}{n+2} + \ldots + \frac{{nC_n}}{2n} \]

### Step 1: Understand the Binomial Expansion
The binomial theorem states that:

\[ (1 + x)^n = \sum_{k=0}^{n} nC_k \cdot x^k \]

This means that the coefficients \(nC_k\) represent the number of ways to choose \(k\) items from \(n\).

**Hint:** Recall the binomial expansion and how coefficients are derived.

### Step 2: Multiply by \(x^{n-1}\)
We multiply both sides of the binomial expansion by \(x^{n-1}\):

\[ x^{n-1} (1 + x)^n = \sum_{k=0}^{n} nC_k \cdot x^{k + n - 1} \]

This gives us:

\[ x^{n-1} (1 + x)^n = nC_0 \cdot x^{n-1} + nC_1 \cdot x^n + nC_2 \cdot x^{n+1} + \ldots + nC_n \cdot x^{2n-1} \]

**Hint:** Understand how multiplying by \(x^{n-1}\) shifts the powers of \(x\).

### Step 3: Integrate Both Sides
Now, we integrate both sides with respect to \(x\):

\[ \int x^{n-1} (1 + x)^n \, dx = \int \left(nC_0 \cdot x^{n-1} + nC_1 \cdot x^n + nC_2 \cdot x^{n+1} + \ldots + nC_n \cdot x^{2n-1}\right) \, dx \]

**Hint:** Remember that integration will increase the power of \(x\) by 1 and divide by the new power.

### Step 4: Evaluate the Integral
We evaluate the left-hand side from 0 to 1:

\[ \int_0^1 x^{n-1} (1 + x)^n \, dx \]

And for the right-hand side, we evaluate each term:

\[ \int_0^1 nC_k \cdot x^{k+n-1} \, dx = nC_k \cdot \frac{1}{k+n} \]

So, the right-hand side becomes:

\[ \sum_{k=0}^{n} nC_k \cdot \frac{1}{k+n} \]

**Hint:** Be careful with the limits of integration and how they affect the terms.

### Step 5: Combine Results
After integrating, we find that:

\[ \int_0^1 x^{n-1} (1 + x)^n \, dx = \sum_{k=0}^{n} \frac{nC_k}{k+n} \]

This matches the original expression we started with.

**Hint:** Make sure to check that both sides of the equation are equal after integration.

### Conclusion
Thus, the value of the given expression is equal to:

\[ \int_0^1 x^{n-1} (1 + x)^n \, dx \]

Upon checking the options, we find that the correct answer is:

**(a)** \(\int_0^1 x^{n-1} (1+x)^n \, dx\)