Question

求数列前N项和

An = n -1,求Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2......+(-1)^(n+1) * bn^2

Answer 1

此题要点是平方和公式1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6，
Tn = (a1)^2 -(a2)^2+(a3)^2-(a4)^2+......+(-1)^(n+1) * an^2
= 0-1^2+2^2-3^2+…+(-1)^(n-1)*(n-1)^2
令Sn=1^2+2^2+3^2+…+(n-1)^2,则
Tn+Sn=0-1^2+2^2-3^2+…+(-1)^(n-1)*(n-1)^2
+1^2+2^2+3^2+…+(n-1)^2
若n为奇数，则
Tn+Sn=2*[2^2+4^2+…+(n-1)^2]
=8*[1^2+2^2+3^2+…+((n-1)/2)^2]
=8*((n-1)/2)*[((n-1)/2)+1]*[2((n-1)/2)+1]/6
=n(n+1)(n-1)/3
Tn=n(n+1)(n-1)/3-Sn
=n(n+1)(n-1)/3-(n-1)n(2(n-1)+1)/6
=n(n-1)/2
若n为偶数，则
Tn+Sn=2*[2^2+4^2+…+(n-2)^2]
=8*[1^2+2^2+3^2+…+((n-2)/2)^2]
=8*((n-2)/2)*[((n-2)/2)+1]*[2((n-2)/2)+1]/6
=n(n-1)(n-2)/3
Tn=n(n-1)(n-2)/3-Sn
=n(n-1)(n-2)/3-(n-1)n(2(n-1)+1)/6
=-n(n-1)/2
从而，
Tn=(-1)^(n-1)*n(n-1)/2.
如果需要求Tn的前n项和Ln，则有
Tn=(-1)^(n-1)*(n^2-n)/2
于是
Ln=(n^2-n)/2-(n^2-n)/2+(n^2-n)/2-(n^2-n)/2+…+(-1)^(n-1)*(n^2-n)/2
=(-1/2)*[0-1^2+2^2-3^2+…+(-1)^n*n^2]-(1/2)*[1-2+3-4+…+(-1)^(n-1)*n]
=(-1/2)*T(n+1)-(1/2)*[1-2+3-4+…+(-1)^(n-1)*n]
若n为奇数，则
Ln=(-1/2)*[-(n+1)n/2]-(1/2)[(n+1)/2]
=(n+1)(n-1)/4
若n为偶数，则
Ln=(-1/2)*[(n+1)n/2]-(1/2)[-n/2]
=-(n^2)/4

Answer 2

这是等差，d=1
n是偶数
Tn=(a2-a1)(a2+a1)+……+[an-a(n-1)][an+a(n-1)]
d=1
所以Tn=a1+……+an=(0+n-1)n/2=(n-1)n/2

n是奇数
Tn=(a2-a1)(a2+a1)+……+[a(n-1)-a(n-2)][a(n-1)+a(n-2)]+an
d=1
所以Tn=a1+……+an=(0+n-1)n/2=(n-1)n/2

综上
Tn=n(n-1)/2

Answer 3

求数列的前n项和