How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$

Question

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?

Answer 1

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.

Answer 2

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 $n^4,$ $$5a_5=16$$

Answer 3

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 ${100 = 33211999680 }$