Binomial Coefficients and Fractions

The sum can be written as : $\displaystyle \sum_{n=1}^7{\dfrac{n}{4} \dbinom{n+3}{3}} = \dfrac{1}{4!}\sum_{n=1}^7{n(n+1)(n+2)(n+3)}$

$\displaystyle= \dfrac{1}{5!}\sum_{n=1}^7{n(n+1)(n+2)(n+3)(n+4) - (n-1)(n)(n+1)(n+2)(n+3)}$

$\displaystyle \text{which telescopes to} \dfrac{1}{5!}7\times 8 \times 9\times 10 \times 11 = \dbinom{11}{5}$

Similar solution with @Kartik Sharma 's

$\begin{aligned} S & = 1 + \frac 12 \binom 53 + \frac 34 \binom 63 + \binom 73 + \frac 54 \binom 83 + \frac 32 \binom 93 + \frac 74 \binom {10}3 \\ & = \frac 14 \binom 43 + \frac 24 \binom 53 + \frac 34 \binom 63 + \frac 44 \binom 73 + \frac 54 \binom 83 + \frac 64 \binom 93 + \frac 74 \binom {10}3 \end{aligned}$

$\begin{aligned} \ \ \ & = \sum_{k=1}^7 \frac k4 \binom {k+3}3 \\ & = \sum_{k=1}^7 \frac {k(k+1)(k+2)(k+3)}{4\cdot 3!} \\ & = \sum_{k=1}^7 \binom {k+3}4 & \small \color{#3D99F6} \text{By Pascal formula: } \binom nk = \binom {n-1}{k-1} + \binom {n-1}k \\ & = \sum_{k=1}^7 \left(\binom {k+4}5 - \binom {k+3}5 \right) \\ & = \sum_{k=1}^7 \binom {k+4}5 - \sum_{k=1}^6 \binom {k+4}5 \\ & = \binom {11}5 \end{aligned}$

Therefore, $a+b = 11 + 5 = \boxed{16}$ .

Extreme 'Stupid' method, brute force out the L.H.S yields $462= 2 \times 3 \times 7 \times 11$ , so $a$ must bigger than or equal to $11$ . Trial and error yields $C^{11}_5$