Question

求一个数列的前n项和

通项公式 a=(n+1)^2

要写过程

Answer 1

法一：
a(n)=(n+1)n + n+1=[n(n+1)(n+2)-(n-1)n(n+1)]/3 + [n(n+1)-(n-1)n]/2 + [(n+1) - n]
s(n)=[1*2*3-0 + 2*3*4-1*2*3 + ... + (n-1)n(n+1)-(n-2)(n-1)n + n(n+1)(n+2)-(n-1)n(n+1)]/3 +
+[1*2-0 + 2*3-1*2 + ... + (n-1)n-(n-2)(n-1) + n(n+1)-(n-1)n]/2 +
+[2-1 + 3-2 + ... + n-(n-1) + (n+1)-n]
=n(n+1)(n+2)/3 + n(n+1)/2 + (n+1)-1
=n[2(n+1)(n+2)+3(n+1) + 6]/6
=n[(n+1)(2n+2+3)+6]/6
=n[(n+1)(2n+5)+6]/6
法二：
a(n)=(n+1)^2=n^2+2n+1,
s(n)=n(n+1)(2n+1)/6 + n(n+1) + n = n(n+1)(2n+1+6)/6 + n = n(n+1)(2n+7)/6 + n

Answer 2

a=(n+1)^2=n^2+2n+1

n^2------n(n+1)(2n+1)/6
2n-------n(n+1)
1--------n
相加就是结果