An algebra problem by Nivedit Jain

$t(n) = (n-4)(n-3)(n-2)(n-1)(n)(n+1)(n+2)$

$t(n) = \frac{(n-4)(n-3)(n-2)(n-1)(n)(n+1)(n+2)(8)}{8}$

$t(n) = \frac{(n-4)(n-3)(n-2)(n-1)(n)(n+1)(n+2)((n+3)-(n-5))}{8}$

Say $f(n) = (n-5)(n-4)(n-3)(n-2)(n-1)(n)(n+1)(n+2)$ and $f(n+1) = (n-4)(n-3)(n-2)(n-1)(n)(n+1)(n+2)(n+3)$

Then $t(n)$ can be written as: $t(n)=\frac{f(n+1)-f(n)}{8}$

Now evaluating the sum, we get $t(5) + t(6) + t(7) + t(8) + t(9) + ..... + t(23) = r=\frac{f(24)-f(5)}{8}$ ,

Now $f(5)=0$ and $f(24)=26*25*24*23*22*21*20*19$ So $r/{7!}=\frac{26*25*24*23*22*21*20*19}{8*7*6*5*4*3*2*1}=1562275$

$\large t(n)=7!^{n+2}C_{n-5}$

$\large \sum_{n=5}^{23} t(n) = \alpha \times 7! = 7! \sum_{n=5}^{23} \binom{n+2}{n-5} = 7!\binom{26}{18}$

We write t n = f(n)*7!. The coefficients can be calculated from the relation: (n - 5)! * t n = (n + 2)!. Then f(n + 1) = f(n)*(n + 3)/(n - 4). Then sum from n = 5 to n =22. f(5) = 1. answer is 1562275. Ed Gray

$7!*t(5)=7!,......7!*t(6)=8!/1!,......7!*t(7)=9!/2!, . . .\\ \therefore~t(n)=\dfrac{(n+2)!}{(n-5)!*7!}.\\ \displaystyle \sum_{n=5}^{23} t(n)=1562275.$