The given series is a combination of two arithmetic progressions. The first progression has a common difference of 2 and starts with 1, and the second progression has a common difference of 2 and starts with 3.


The formula for the sum of n terms of an arithmetic progression is:

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

Where Sn is the sum of n terms, a is the first term, d is the common difference, and n is the number of terms.


The first progression starts with 1 and has a common difference of 2. Therefore, the formula for the sum of n terms of the first progression is:

Sn1 = n/2 * [2(1) + (n-1)2]
= n/2 * [2 + 2n - 2]
= n/2 * (2n)
= n^2


The second progression starts with 3 and has a common difference of 2. Therefore, the formula for the sum of n terms of the second progression is:

Sn2 = n/2 * [2(3) + (n-1)2]
= n/2 * [6 + 2n - 2]
= n/2 * (2n + 4)
= n(n+1)


To calculate the sum of the entire series, we need to add the sum of the first progression and the sum of the second progression:

S = Sn1 + Sn2
= n^2 + n(n+1)
= n^2 + n^2 + n
= 2n^2 + n

Therefore, the sum of the series 1.3.5.7, 3.5.7.9 till n is 2n^2 + n.

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