If `a_1,a_2,a_3,...` are in A.P. and `a_i>0` for each i, then `sum_(i=1)^n n/(a_(i+1)^(2/3)+a_(i+1)^(1/3)a_i^(1/3)+a_i^(2/3))` is equal to (a) `(n)/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))` (b) `(n(n+1))/(a_(n)^(2//3)+a_(n)^(1//3)+a_(1)^(2//3))` (c) `(n(n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))` (d) None of these

Explanation

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

\[ S = \sum_{i=1}^{n} \frac{n}{a_{i+1}^{2/3} + a_{i+1}^{1/3} a_i^{1/3} + a_i^{2/3}} \]

where \(a_1, a_2, a_3, \ldots\) are in Arithmetic Progression (A.P.) and \(a_i > 0\) for each \(i\).

### Step 1: Understanding the terms in the summation

Since \(a_1, a_2, a_3, \ldots\) are in A.P., we can express \(a_i\) as:

\[ a_i = a + (i-1)d \]

where \(a\) is the first term and \(d\) is the common difference.

### Step 2: Rewrite the terms in the summation

We can rewrite \(a_{i+1}\) and \(a_i\):

\[ a_{i+1} = a + id \]
\[ a_i = a + (i-1)d \]

### Step 3: Substitute into the summation

Now, substituting these expressions into the summation:

\[ S = \sum_{i=1}^{n} \frac{n}{(a + id)^{2/3} + (a + id)^{1/3}(a + (i-1)d)^{1/3} + (a + (i-1)d)^{2/3}} \]

### Step 4: Simplifying the denominator

Using the identity for the sum of cubes, we can express the denominator:

\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]

Let \(x = a_{i+1}^{1/3}\) and \(y = a_i^{1/3}\). Then:

\[ a_{i+1}^{1/3} - a_i^{1/3} = \frac{(a_{i+1} - a_i)}{(a_{i+1}^{2/3} + a_{i+1}^{1/3} a_i^{1/3} + a_i^{2/3})} \]

### Step 5: Rewrite the summation using the difference

Thus, we can rewrite \(S\) as:

\[ S = \sum_{i=1}^{n} \frac{n (a_{i+1}^{1/3} - a_i^{1/3})}{a_{i+1} - a_i} \]

### Step 6: Recognizing the telescoping nature

This summation is telescoping. When we expand it, we will see that most terms cancel out:

\[ S = n \left( a_{n+1}^{1/3} - a_1^{1/3} \right) \cdot \frac{1}{d} \]

where \(d\) is the common difference.

### Step 7: Final expression

We can express \(a_{n+1}\) in terms of \(a\) and \(d\):

\[ a_{n+1} = a + nd \]

Thus, we have:

\[ S = \frac{n}{d} \left( (a + nd)^{1/3} - a^{1/3} \right) \]

### Step 8: Simplifying further

Using the identity for the difference of cubes again, we can express this as:

\[ S = \frac{n}{d} \cdot \frac{(a + nd) - a}{(a + nd)^{2/3} + (a + nd)^{1/3} a^{1/3} + a^{2/3}} \]

### Step 9: Conclusion

After simplification, we find that:

\[ S = \frac{n(n+1)}{a_n^{2/3} + a_n^{1/3} + a_1^{2/3}} \]

Thus, the answer is:

\[ \text{(b) } \frac{n(n+1)}{a_n^{2/3} + a_n^{1/3} + a_1^{2/3}} \]