How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?

Hint:
The $n$th term $T(n)$ is $$(2n-1)(2n)(2n+1)(2n+2) =16n^4+16n^3-4n^2-4n$$
WLOG $$T(n)=P(n+1)-P(n)$$ where $P(m)=\sum_{r=0}^ua_rm^r$
$$\sum_{r=0}^5a_r((n+1)^r-n^r)$$
$$=a_1+a_2(2n+1)+a_3(3n^3+3n+1)+a_4(4n^3+6n^2+4n+1)a+a_5(5n^4+10n^3+10n^2+5n+1)+\cdots$$
Clearly, $a_r=0$ for $r\ge6$
Now compare the  coefficients of $n^4,n^4,n^2,n$ and constants to find $a_j;0\le j\le5$
Comparing the coefficients of $n^4,$ $$5a_5=16$$
