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 can break down each term.

The general term of the given series can be written as:

Term(n) = (2n - 1) / (n(n + 1)(n + 2))

Now, let's find the pattern in the given series:

1st term (n=1) = (2*1-1) / (1*(1+1)*(1+2)) = 1/6
2nd term (n=2) = (2*2-1) / (2*(2+1)*(2+2)) = 3/24 = 1/8
3rd term (n=3) = (2*3-1) / (3*(3+1)*(3+2)) = 5/60 = 1/12
4th term (n=4) = (2*4-1) / (4*(4+1)*(4+2)) = 7/120

From the above pattern, we can observe that the numerator of each term is an odd number, starting from 1 and increasing by 2 with every term.

Now, let's try to find a general formula for the series, Sn, to calculate the sum to n terms.

Sn = (1/6) + (1/8) + (1/12) + (1/20) + ... + (2n-1) / (n*(n+1)*(n+2))

To simplify the series, we can express each term as a common fraction with the least common multiple (LCM) of (n)(n+1)(n+2) as the denominator.

Sn = [(2n-1) * (n+1) * (n+2) + (n) * (n+2) + (n) * (n+1)] / (6n)

Simplifying the numerator:

Sn = [2n^3 + 6n^2 + 3n + 2n^2 + 4n + 2n] / (6n)
= [2n^3 + 8n^2 + 9n] / (6n)
= (n^2 + 4n + 3n) / (3n)

Sn = (n(n+3)) / (3n)
= (n+3)/3

Therefore, the sum of the given series to n terms is given by:

Sn = (n+3)/3