Question

给定数列1,2+3+4,5+6+7+8+9,10+11+12+13+14+15+16,...求该数列的通项公式．（详细）

Answer 1

S<1>=1，S<2>=1+2+3+4，S<3>=1+2+3+…+9，
故S<n>=1+2+…+n²=n²(1+n²)/2
a<n>=S<n>-S<n-1>=[n²(1+n²)/2]-(n-1)²[1+(n-1)²]/2
=2n³-3n²+3n-1

Answer 2

根据规律,可以得出,第n项共有2n-1项.
所以前面共有1+3+5+…+（2n-3)=(n-1)^2=n^2-2n+1
所以an=(n^2-2n+2)+(n^2-2n+3)+(n^2-2n+4)+……+(n^2-2n+2n)
=(2n-1)(n^2-2n+2+n^2)/2=2n^3-3n^2+3n-1.