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?
 A: 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 $n^4,$ $$5a_5=16$$
A: I don't have reputation for comment. so trying to answer here.
${S_n= (2n-1)(2n)(2n+1)(2n+2)}$, simplifying it.

*

*${S_n = (2n) \cdot (4n^2-1) \cdot (2n+2)}$,

*${S_1+S_2+S_3+S_4= 24+360+1680+5040=7104}$

*Simple program gives sum till ${100 = 33211999680 }$
A: General term is:
$$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$
Now we give values 1, 3,... n :
$1\times2\times3\times4=1^4+6\times1^3+11\times 1^2+6\times 1$
$3\times4\times5\times6=3^4+6\times3^3+11\times 3^2+6\times 3$
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$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$
Now we sum up:
$A=S_4+6S_3+11S_2+6S_1$
We use  well known formulas for $S_n$ for odd numbers, for example:
$S_3=1^3+3^3+ . . . +n^3=n^2(2n^2-1)$
Here $n= 96$, we plug this in and get the sum.
