Question

(1) Sn=2+3+5+8+12+17+23+30+38+47+57 能否用以公式表达？

Answer 1

A2 - A1 = 1
A3 - A2 = 2
A4 - A3 = 3
A5 - A4 = 4
…………
An - A(n-1) = n-1
左、右两边同时相加，可以得到
An - A1 = 1 + 2 + 3 + …… + (n-1) = n(n-1)/2
An = 2 + n(n-1)/2 = (n²-n+4)/2
所以，
Sn = 1/2 * ∑(n²-n+4)
= 1/2 * [∑n² - ∑n + ∑4]
= 1/2 * [n(n+1)(2n+1)/6 - n(n+1)/2 + 4n]
= 1/2 * 1/6 * [n(n+1)(2n+1) - 3n(n+1) + 24n]
= n/12 * [(n+1)(2n+1) - 3(n+1) + 24]
= n/12 * [(n+1)(2n+1-3) + 24]
= n/12 * [2(n+1)(n-1) + 24]
= n/12 * 2 * [(n+1)(n-1) + 12]
= n/6 * [n² - 1 +12]
= n(n²+11)/6
把 n = 11 代入上面的求和公式：
Sn = 11×(11²+11)/6
= 11×11×12/6
= 242

Answer 2

解，a1=2，
an=a(n-1)+(n-1)
a(n-1)=a(n-2)+(n-2)
，，，，，，，，，，，，
，，，，，，，，，，，，
则an=(n-1)+(n-2)+，，，2+1+a1
=(n-1)n/2+2
=n^2/2-n/2+2
则Sn=a1+a2+a3+，，，+αn
=1/2(1+2^2+，，+n^2)-1/2(1+2+，，n)+2n
=n(n+1)(2n+1)/12-(n+1)n/4+2n

追问

感谢你的解析 我是在中学课本看到的 没想到这么复杂 谢谢你了