Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$ 
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?

I change a little the notations from your original wording.
$T(n)=\dfrac{n(n+1)}{2}$
$S(n)=\sum\limits_{k=1}^nT(k)^2$
Cheating on the resulting formula for $f(n)=\dfrac{n(n+1)(n+2)(3n^2+6n+1)}{60}$ 
we notice that $f(0)=f(-1)=f(-2)=0$. 
Is there a way to exploit these negative indices in order to find the $(3n^2+6n+1)$ part more easily?
In fact there is, $T(n)$ is perfectly defined for negative numbers so this part does not pose any issue. 
But how to interpret $S(-n)$?
To be consistent with the summation identity $\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$ when $a<b<c\ $ one need to set $\ \sum\limits_M^m=-\sum\limits_{m+1}^{M-1}$ for the case $m<M$. 
This results in $\displaystyle S(-n)=\sum\limits_{k=1}^{-n}T(k)^2=-\sum\limits_{k=-(n-1)}^0T(k)^2$
I let you do the calculation to verify that $\begin{array}{lcr}S(0)&=&0\\S(-1)&=&0\\ S(-2)&=&0\\ S(-3)&=&-1\\S(-4)&=&-10\\S(-5)&=&-46\end{array}$
Now by applying the identification method (I describe in my other post) to $$S(n)=a_0+a_1\,n+a_2\,n(n+1)+a_3\,n(n+1)(n+2)+\cdots+a_5\,n(n+1)(n+2)(n+3)(n+4)$$
You immediately get $a_0=a_1=a_2=0$
The others coefficients give you $$S(n)=n(n+1)(n+2)\bigg[\frac 16-\frac 14(n+3)+\frac 1{20}(n+3)(n+4)\bigg]=n(n+1)(n+2)\left(\dfrac{3n^2+6n+1}{60}\right)$$
