Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6)
To find the sum of the given series, we need to determine the pattern of the terms. Upon observing the terms carefully, we can see that the numerator of each term is an odd number starting from 1, and the denominator consists of consecutive integers starting from 1.
The general term of the series can be written as follows:
Term(n) = (2n - 1) / (n * (n + 1) * (n + 2)), where n represents the position of the term.
Now, to find the sum to 'n' terms of the given series, we can proceed as follows:
1. Start by substituting each term of the series into the general term expression:
S(n) = (2*1 - 1) / (1 * (1 + 1) * (1 + 2)) + (2*2 - 1) / (2 * (2 + 1) * (2 + 2)) + ... + (2n - 1) / (n * (n + 1) * (n + 2))
2. Simplify the expression:
S(n) = 1 / (1 * 2 * 3) + 3 / (2 * 3 * 4) + 5 / (3 * 4 * 5) + 7 / (4 * 5 * 6) + ... + (2n - 1) / (n * (n + 1) * (n + 2))
3. Combine the fractions with a common denominator:
S(n) = [(2n - 1) * (n + 1) * (n + 2) + 3 * n * (n + 2) * (n + 3) + 5 * (n + 2) * (n + 3) * (n + 4) + 7 * (n + 3) * (n + 4) * (n + 5) + ...] / (n * (n + 1) * (n + 2) * (n + 3) * (n + 4) * ... * (n + n))
4. Simplify the numerator expression:
S(n) = [(2n^3 + 9n^2 + 13n + 6) + (n^3 + 6n^2 + 11n + 6) + (n^3 + 7n^2 + 18n + 15) + (n^3 + 6n^2 + 17n + 14) + ...] / (n!)
5. Combine like terms within the numerator:
S(n) = (2n^3 + 2n^2 + 4n + 1) / (n * (n + 1) * (n + 2) * (n + 3) * (n + 4) * ... * (n + n))
Therefore, the sum to 'n' terms of the given series is:
S(n) = (2n^3 + 2n^2 + 4n + 1) / (n!)