Question

一道高中数列题 数列{n(n+1)(n+2)(n+3)}的前n项和为 请写出具体过程谢谢！

Answer 1

舍予9217错了哈,
a(n)=(n+3)!/(n-1)!
而不是a(n)=(n+3)! - (n-1)!
显然, a(1)=4! 不等于 4!- 0!=4!-1
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a(n)= n(n+1)(n+2)(n+3) = (1/5)[n(n+1)(n+2)(n+3)(n+4) - (n-1)n(n+1)(n+2)(n+3) ]
a(n-1)=(1/5) [ (n-1)n(n+1)(n+2)(n+3) - (n-2)(n-1)n(n+1)(n+2)]
...
a(2) = (1/5) [2*3*4*5*6 - 1*2*3*4*5]
a(1) = (1/5) [ 1*2*3*4*5 - 0]

s(n)=a(1)+a(2)+...+a(n)=(1/5) [n(n+1)(n+2)(n+3)(n+4) - 0] = n(n+1)(n+2)(n+3)(n+4)/5

Answer 2

an=(n+3)!-(n-1)!
a(n-1)=(n+2)!-(n-2)!
a(n-2)=(n+1)!-(n-3)!
~~~~~~~~~~~
a4=7!-3!
a3=6!-2!
a2=5!-1!
a1=4!
将上式两边同时相加得
sn=（4！+5！+6！+~~+（n+2）！）-（1！+2！+3！+~~+（n-2）!）
sn=(n+2)!+(n+1)!+n!-1!-2!-3!

Answer 3

2n!/(n-1)!