Let `T_r` be the rth term of an A.P. whose first term is -1/2 and common difference is 1, then `sum_(r=1)^n sqrt(1+ T_r T_(r+1) T_(r+2) T_(r+3))`

Options

A. `(n(n+1)(2n +1))/(6 )-(5n)/(4)`
B. `(n(n+1)(2n +1))/(6 )-(5n)/(4)+1/4`
C. `(n(n+1)(2n +1))/(6 )-(5n)/(4)+1/2`
D. `(n(n+1)(2n +1))/(12)-(5n)/(8)+1`

Answer

To solve the problem, we need to find the sum

\[ \sum_{r=1}^{n} \sqrt{1 + T_r T_{r+1} T_{r+2} T_{r+3}} \]

where \( T_r \) is the r-th term of an arithmetic progression (A.P.) with the first term \( A = -\frac{1}{2} \) and common difference \( D = 1 \).

### Step 1: Determine the general term \( T_r \)

The r-th term of an A.P. is given by the formula:

\[ T_r = A + (r-1)D \]

Substituting the values of \( A \) and \( D \):

\[ T_r = -\frac{1}{2} + (r-1) \cdot 1 = r - \frac{3}{2} \]

### Step 2: Write the terms \( T_r, T_{r+1}, T_{r+2}, T_{r+3} \)

Now, we can express the next three terms:

- \( T_{r+1} = -\frac{1}{2} + r \cdot 1 = r - \frac{1}{2} \)
- \( T_{r+2} = -\frac{1}{2} + (r+1) \cdot 1 = r + \frac{1}{2} \)
- \( T_{r+3} = -\frac{1}{2} + (r+2) \cdot 1 = r + \frac{3}{2} \)

### Step 3: Calculate the product \( T_r T_{r+1} T_{r+2} T_{r+3} \)

Now we need to compute the product:

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

This can be simplified using the difference of squares:

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

Calculating each part:

\[ (r - \frac{3}{2})(r + \frac{3}{2}) = r^2 - \left(\frac{3}{2}\right)^2 = r^2 - \frac{9}{4} \]

\[ (r - \frac{1}{2})(r + \frac{1}{2}) = r^2 - \left(\frac{1}{2}\right)^2 = r^2 - \frac{1}{4} \]

Now, multiplying these results:

\[ T_r T_{r+1} T_{r+2} T_{r+3} = \left(r^2 - \frac{9}{4}\right)\left(r^2 - \frac{1}{4}\right) \]

### Step 4: Expand the product

Using the distributive property:

\[ = r^4 - \frac{1}{4}r^2 - \frac{9}{4}r^2 + \frac{9}{16} \]

Combining like terms:

\[ = r^4 - \frac{10}{4}r^2 + \frac{9}{16} = r^4 - \frac{5}{2}r^2 + \frac{9}{16} \]

### Step 5: Substitute into the sum

Now we can substitute this back into our original sum:

\[ \sum_{r=1}^{n} \sqrt{1 + T_r T_{r+1} T_{r+2} T_{r+3}} = \sum_{r=1}^{n} \sqrt{1 + r^4 - \frac{5}{2}r^2 + \frac{9}{16}} \]

This simplifies to:

\[ = \sum_{r=1}^{n} \sqrt{r^4 - \frac{5}{2}r^2 + \frac{25}{16}} \]

### Step 6: Further simplification

Recognizing that:

\[ r^4 - \frac{5}{2}r^2 + \frac{25}{16} = \left(r^2 - \frac{5}{4}\right)^2 \]

Thus, we have:

\[ \sqrt{1 + T_r T_{r+1} T_{r+2} T_{r+3}} = \left| r^2 - \frac{5}{4} \right| \]

### Step 7: Final sum calculation

Now we can evaluate the sum:

\[ \sum_{r=1}^{n} \left| r^2 - \frac{5}{4} \right| \]

This will depend on the values of \( r \) and whether \( r^2 \) is greater or less than \( \frac{5}{4} \).