ตอนที่3 ข้อ5
$\sum_{n=1}^{100} \frac{n^2+3n +1}{(n+1)!(n+2)!}$ = $\sum_{n=1}^{100} \frac{n^2+2n}{(n+1)!(n+2)!}$ + $\sum_{n=1}^{100} \frac{n +1}{(n+1)!(n+2)!}$
= $\sum_{n=1}^{100} \frac{n}{(n+1)!(n+1)!}$ + ($\sum_{n=1}^{100} \frac{n +2}{(n+1)!(n+2)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+2)!}$)
=($\sum_{n=1}^{100} \frac{n +1}{(n+1)!(n+1)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+1)!}$) + ($\sum_{n=1}^{100} \frac{n +2}{(n+1)!(n+2)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+2)!}$)
=$\sum_{n=1}^{100} \frac{n +1}{(n+1)!(n+1)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+1)!}$ + $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+1)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+2)!}$
=$\sum_{n=1}^{100} \frac{1}{n!(n+1)!}$ - $\sum_{n=1}^{100} \frac{1}{(n+1)!(n+2)!}$
= $\frac{1}{1!2!}$ - $\frac{1}{101!102!}$
เมื่อ $\frac{1}{1!2!}$ - $\frac{1}{101!102!}$ = $\frac{(a!)^2\times{51} -1}{102(a!)^2}$
=$\frac{1}{2}$ - $\frac{1}{102(a!)^2}$
ดังนั้น $\frac{1}{101!101!}$ = $\frac{1}{(a!)^2}$
$a=101$
เพิ่งฝึกพิมลาเท็กเป็นข้อแรก ถ้าผิดตรงไหนขออภัยด้วยนะคะ
