The sum of the following series $1+6+\dfrac{9(1^2+2^2+3^2)}{7}+\dfrac{12(1^2+2^2+3^2+4^2)}{9}+\dfrac{15(1^2+2^2+....+5^2)}{11}+.....$ up to $15$ terms is:

Observe that the third term onwards fits the pattern where the nth term is given by 3(n-1) multiplied by the sum of the first (n-1) squares, divided by (2n+1). Using the sum of squares formula n(n+1)(2n+1)/6, the expression simplifies to (n-1)(n)(n)(n-1)/2, which further reduces to (n-1)^2 * n^2 / 2. The 15-term series includes the constants 1 and 6 for the first two terms, so substituting n = 3 to 15 into the formula and summing the results gives 7820.

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