Question

(1)已知数列{an}中,an=2n-3+2^n,求数列{an}的前n项和Sn,(2)求和:Sn=-1+3-5+7…+(-1)^n(2n-1)

Answer 1

解：1、
Sn=a1+a2+....+an
=(2*1-3+2^1)+(2*2-3+2^2)+.......+(2n-3+2^n)
=2(1+2+.....+n)-3n+(2^1+2^2+.....+2^n)
=n(n+1)-3n+2^(n+1)-2
=n^2-2n-2+2^(n+1)
2、当n为奇数时
Sn=-1+3-5+7…+(-1)^n(2n-1)
Sn=-1-3-5-7-.....-(2n-1)+2(3+7+11+.......+(4n-1))
自已算吧
当n为偶数时也自已算吧

Answer 2

1
Sn = S{2n} + S{-3} + S{2^n}
= 2*(1+n)*n/2 + (-3)*n + (2-2^(n+1))/(1-2)
= n² +n -3n + 2^(n+1) -2
= n ²- 2n + 2^(n+1) -2

2
如为偶数项
Sn = (-1+3)+(-5+7)…+　(-2(n-1)-1 + 2n-1) = 2 * (n/2) = n
如为奇数项

Sn = (-1+3)+(-5+7)…+　(-2(n-2)-1 + 2(n-1)-1) - (2n-1) = 2* (n-1)/2 - (2n-1) = -n

Answer 3

（1）Sn=(2+4+6+……+2n)-3n+(2+2^2+2^3+……+2^n)=n(n+1)-3n+2^(n+1）-2=2^(n+1）+n^2-2n-2
(2)当n是偶数时，Sn=n;当n时奇数时，Sn=(n-1)+(-1)^n(2n-1)