Animated Solution for Mathematics - Sequence and Series: The sum of the following series $1+6+\frac{9(1^2+2^2+3^2)}{7}+\frac{12(1^2+2^2+3^2+4^2)}{9}+\frac{15(1^2+2^2+...+5^2)}{11}+...$ up to 15 terms, is:

Welcome, warriors of JEE! Today, we are going to dismantle a series that, at first glance, looks like a chaotic mess of numbers and fractions. But remember, in the world of competitive mathematics, chaos is just order waiting to be discovered.

Our series is $1+6+\frac{9(1^2+2^2+3^2)}{7}+\frac{12(1^2+2^2+3^2+4^2)}{9}+\dots$. The goal is to find the sum of the first $15$ terms, $S_{15}$.

The first step is to stop being intimidated. Look at the 3rd and 4th terms; they have a clear, repeating structure: a multiplier, a sum of squares, and a denominator.

We can rewrite $1$ as $\frac{3(1^2)}{3}$ and $6$ as $\frac{6(1^2+2^2)}{5}$. Now, the entire series is a family.

Now that we have unmasked the pattern, let us find its DNA. We need the general term $T_n$.

Look at the multipliers in the numerator: $3,6,9,12,\dots$. This is an Arithmetic Progression (A.P.) with first term $a=3$ and common difference $d=3$. The $n$-th term is simply $3n$.

Now, look at the denominators: $3,5,7,9,\dots$. This is another A.P. with $a=3$ and $d=2$. The $n$-th term is $2n+1$.

Finally, the core of each term is the sum of squares: ${\sum }_{k=1}^n{k}^2$. We know the standard identity:

Putting it all together, our general term is:

This is the moment of truth. Let us substitute the sum of squares formula into our expression for $T_n$:

Look closely. The $(2n+1)$ in the numerator and the $(2n+1)$ in the denominator are perfectly set up to vanish. They cancel out, leaving us with:

Simplifying $\frac{3}{6}$ to $\frac{1}{2}$, we get:

This is the beauty of algebra; what looked like a terrifying fraction has collapsed into a simple polynomial.

Now, we calculate $S_{15}={\sum }_{n=1}^{15}{T}_n=\frac{1}{2}[{\sum }_{n=1}^{15}{n}^3+{\sum }_{n=1}^{15}{n}^2]$. We use our standard summation formulas.

For the cubes:

For the squares:

Adding these together, we get $14400+1240=15640$. Finally, we divide by $2$ to get the result.

The final sum is $7820$.