当n为奇数时，an=2n；当n为偶数时，an=2n-1 求前n项和Sn

Question

如题所述

Answer 1

分开来求，n为奇数的时候奇数项有（n-1）/2 +1项，用等差数列求和公式，偶数项有（n-1）/2项求和，两次求和相加即可
n为偶数就简单了啊，奇数项和偶数项都是n/2项啊

Answer 2

an= 2n ; n is odd
=2n-1 ; n is even

S1 = a1= 1
S2 = a1+a2 = 1+2 =3
S3= S2 + a3= 3+5 =8

when n>=4
when n is odd
Sn= a1+a2+...+an
= (a1+a3+...+an) + (a2+a4+...+a(n-1) )
=(a1+an)(n+1)/4 + (a2+a(n-1) )(n-1)/4
= (2n-1+1)(n+1)/4 + [2(n-1)+ 2](n-1)/4
= (1/2)n(n+1) + (1/2)n(n-1)
=n^2

when n is even
Sn= a1+a2+...+an
= (a1+a3+...+a(n-1)) + (a2+a4+...+an )
=(a1+a(n-1))(n-1)/4 + (a2+an )(n-1)/4
= (2(n-1)-1+1)n/4 + [2n+ 2]n/4
= (1/2)n(n-1) + (1/2)n(n+1)
=n^2
ie
Sn = 1 ; n=1
=3 ; n=2
=8 ;n=3
=n^2 ; n>3

Answer 3

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