Question

求解一道数列求和的数学题.求Sn=1∧2-2∧2+3∧2-4∧2+...+（-1)∧(n-1)×n∧2

Answer 1

若n为偶数,则Sn=1^2-2^2+3^2-4^2+……+(-1)^(n+1) n^2

=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+……+((n-1)-n)((n-1)+n)

=-(1+2+3+4+……+n-1+n)

=-(n+1)n/2,

若n为奇数,则Sn=1^2-2^2+3^2-4^2++(-1)^(n+1) n^2

=1+(3-2)(3+2)+(5-4)(5+4)+……+(n-(n-1))(n+(n-1))

=1+2+3+4+……+n-1+n

=(n+1)n/2

综上可知Sn=(-1)^(n+1) (n+1)n/2.