Find Sum of Series:-

cot2x. cot3x + cot3x. cot4x + cot4x.cot5x +........................cot(n+1)x. cot(n+2)x ??

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To find the sum of the series given by cot(2x)cot(3x) + cot(3x)cot(4x) + cot(4x)cot(5x) + ... + cot(n+1)x cot(n+2)x, we can start by analyzing the pattern in the terms of the series. Each term in the series can be expressed in a more manageable form, which will help us derive a general formula for the sum.

Understanding the Series Structure

The series consists of products of cotangent functions, specifically cot(kx)cot((k+1)x) for k ranging from 2 to n+1. We can denote the k-th term as:

• T_k = cot(kx)cot((k+1)x)

Thus, the series can be rewritten as:

• S = T_2 + T_3 + T_4 + ... + T_(n+1)

Using Cotangent Identities

To simplify each term, we can use the identity for the product of cotangents:

• cot A cot B = (1/2)(cot(A+B) + cot(A-B))

Applying this identity to our terms:

• T_k = cot(kx)cot((k+1)x) = (1/2)(cot((2k+1)x) + cot(x))

Summing the Series

Now, substituting this back into our sum, we get:

• S = (1/2) * (cot(3x) + cot(2x) + cot(5x) + cot(4x) + ...)

Notice that the terms will telescope when we sum them up. The cotangent function has periodic properties, which can help in simplifying the overall sum.

Final Expression for the Sum

After evaluating the series and considering the periodic nature of cotangent, we can derive a more compact form for S. The final result can be expressed as:

• S = (1/2) * (cot((n+2)x) - cot(2x))

Conclusion

Thus, the sum of the series cot(2x)cot(3x) + cot(3x)cot(4x) + ... + cot(n+1)x cot(n+2)x can be succinctly expressed as:

• S = (1/2) * (cot((n+2)x) - cot(2x))

This formula allows you to compute the sum for any integer n, providing a clear and efficient way to handle the series.