Combinatorial Sum

We start by writing $r^3 \dbinom nr$ as a linear combination of binomial coefficients . Note that

$r^3 = r(r-1)(r-2) + 3r(r-1) + r$

The summation becomes

$\begin{aligned} && \displaystyle \sum_{r=1}^n \left [ r(r-1)(r-2) + 3r(r-1) + r \right] \dbinom nr \\ &=&\displaystyle \sum_{r=4}^n \left[ r(r-1)(r-2)\dbinom nr \right ] + 3 \sum_{r=3}^n \left [r(r-1)\dbinom nr \right ]+ \sum_{r=2}^n \left [ (r\dbinom nr \right ] \end{aligned}$

Note that

$\begin{aligned} r(r-1)(r-2)\dbinom nr &=& r(r-1)(r-2) \cdot \dfrac{n!}{(n-r)!r!} \\ &=& \dfrac{n!}{(n-r)!(r-3)!} \\ &=& n(n-1)(n-2) \cdot \dfrac{(n-3)!}{(n-r)!(r-3)!} \\ &=& n(n-1)(n-2) \cdot \dbinom {n-3}{r-3} \end{aligned}$

Similarly, $r(r-1)\dbinom nr = n(n-1)\cdot \dbinom{n-2}{r-2}$ and $r\dbinom nr = n\dbinom{n-1}{r-1}$ .

Our sum is equal to

$\displaystyle n(n-1)(n-2) \sum_{r=3}^n \dbinom{n-3}{r-3} + 3n(n-1) \sum_{r=2}^n \dbinom{n-2}{r-2} + n \sum_{r=1}^n \dbinom{n-1}{r-1}.$

Once we apply the property $\displaystyle \sum_{j=0}^n \dbinom nj = 2^j$ , the summation simplifies to

$n(n-1)(n-2) \cdot 2^{n-3} + 3n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} = 2^{n-3} (n^3 + 3n^2 ) .$

Best and the quickest solution is to put value of (r,n) as (1,1), (1,2) and (1,3) (as the given equation is valid for all r and n) and obtain 3 simple equations in a, b and c.

For (r,n)= (1,1): The equation is -> a+b+c=3

For (r,n)= (1,2): The equation is -> 4a+2b+c=12

For (r,n)= (1,3): The equation is -> 9a+3b+c=27

Solving the above equations, we get a=3, b=0 and c=0

Hence, a^ ${2}$ +b^ ${2}$ +c^ ${2}$ = $\boxed{9}$