Question

求数列an=(2n-1)^2的前n项和

Answer 1

an=(2n-1)^2=4n^2-4n+1,
an-1-4(n-1)^2-4(n-1)+1,
……………………
a1=4*1^2-4+1
Sn=4*(1^2+2^2+...+n^2)-4(1+2+...+n)+n
=4*1/6n(n+1)(2n+1)-4[n*(n+1)/2]+n
=2/3n(n+1)(2n+1)-2n*(n+1)+n
=n*(n+1)*[2/3(2n+1)-2]+n
=n*(n+1)*(4/3n-4/3)+n
=4/3n*(n+1)*(n-1)+n
=4/3n*(n^2-1)+n
=4/3n*[(n^2-1)+3/4]
=4/3n*(n^2-1/4)
=1/3n*(4n^2-1)
=1/3n*(2n+1)(2n-1)
=4/3n^3-1/3n
最后两步都是答案吧..我记得好像要全展开的..要不扣分是不是...
二楼的...
数列1^2,2^2,3^2,……,n^2的前n项和1/6n(n+1)(2n+1)
把n换成2n,即前2n项和1/3n(2n+1)(4n+1)
好像不能这么直接用的吧..是多了n项呢啊~不只是二倍的关系...
1/3n(2n+1)(4n+1)只是两倍的1^2,2^2,3^2,……,n^2的和...n项后面的是n+1,不是1喔~