Find the sum of the series
`cot^(-1) 7+cot^(-1)13+cot^(-1)21+cot^(-1)31+...` to n terms

Answer

To find the sum of the series \( \cot^{-1}(7) + \cot^{-1}(13) + \cot^{-1}(21) + \cot^{-1}(31) + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Identify the pattern in the series
The terms in the series can be expressed as:
- First term: \( 7 = 1 + 3 \times 2 \)
- Second term: \( 13 = 1 + 4 \times 3 \)
- Third term: \( 21 = 1 + 5 \times 4 \)
- Fourth term: \( 31 = 1 + 6 \times 5 \)

In general, the \( n \)-th term can be expressed as:
\[ a_n = 1 + (n + 2)(n + 1) \]

### Step 2: Express the \( n \)-th term
The \( n \)-th term can be simplified:
\[ a_n = 1 + (n + 2)(n + 1) = 1 + n^2 + 3n + 2 = n^2 + 3n + 3 \]

### Step 3: Rewrite the series using the formula for cotangent
We can express the series using the identity:
\[ \cot^{-1}(a) + \cot^{-1}(b) = \cot^{-1} \left( \frac{ab - 1}{a + b} \right) \]
However, we can also use the formula:
\[ \cot^{-1}(x) = \cot^{-1}(a) - \cot^{-1}(b) \text{ where } a = 1 + ab \text{ and } b = a - b \]

### Step 4: Apply the formula to the series
Using the identity:
\[ \cot^{-1}(x) = \cot^{-1}(a) - \cot^{-1}(b) \]
for each term, we find:
\[ \cot^{-1}(7) + \cot^{-1}(13) + \cot^{-1}(21) + \ldots + \cot^{-1}(a_n) \]
can be rewritten as:
\[ \cot^{-1}(3) - \cot^{-1}(2) + \cot^{-1}(4) - \cot^{-1}(3) + \ldots + \cot^{-1}(n + 2) - \cot^{-1}(n + 1) \]

### Step 5: Simplify the series
Notice that this is a telescoping series:
\[ \cot^{-1}(n + 2) - \cot^{-1}(2) \]
After cancellation, we are left with:
\[ \cot^{-1}(n + 2) - \cot^{-1}(2) \]

### Step 6: Final expression
Thus, the sum of the series up to \( n \) terms is:
\[ \cot^{-1}(n + 2) - \cot^{-1}(2) \]

### Conclusion
The final result for the sum of the series \( \cot^{-1}(7) + \cot^{-1}(13) + \cot^{-1}(21) + \ldots \) up to \( n \) terms is:
\[ \cot^{-1}(n + 2) - \cot^{-1}(2) \]