已知（an）满足an=(2n-1)*2的n次方，求（an)前n项和为Sn.(已知an =(2n-1) 2^n,求前n项和)
已知an =(2n-1) 2^n,求前n项和
a1=1*2^1
a2=3*2^2
a3=5*2^3
...
an=(2n-1)*2^n
Sn=a1+a2+a3+...+an=1*2^1+3*2^2+5*2^3+...+(2n-1)*2^n
Sn/2=1+3*2^1+5*2^2+...+(2n-1)^2(n-1)
Sn-(Sn/2)=-[1+2*2^1+2*2^2+2*2^3+...2^2(n-1)]+(2n-1)*2^n
=-1-{2*2^1[1-2^(n-1)]/(1-2)}+(2n-1)*2^n
=-1+4-2*2^n+(2n-1)*2^n
Sn/2 =(2n-3)*2^n +3
Sn=(2n-3)*2^(n+1) +6
已知数列an=(2n-1)/2^n,求它的前n项和
an=(2n-1)/2^n =2n/2^n-1/2^n 设 Cn=2/2^n=0.5^(n-1) bn=2n/2^n=n*0.5^(n-1)=n*Cn Cn为首项为1 公比为0.5的等比数列 Ck+....+Cn =Ck*[1-0.5^(n-k+1)]/(1-0.5) =C1*0.5^(k-1)*[1-0.5^(n-k+1)]/(1-0.5) =0.5^(k-2)-0.5^(n-1) =4*0.5^k-0.5^(n-1) 数列bn的前n项和Tn Tn =b1+b2+b3+....+bn =C1+2*C2+3*C3+....+n*Cn =（C1+C2+C3+....+Cn）+（C2+C3+....+Cn）+（C3+....+Cn）+...+（Cn-1+Cn）+Cn =4*0.5^1-0.5^(n-1) +4*0.5^2-0.5^(n-1) +4*0.5^3-0.5^(n-1) +.... +4*0.5^袱筏递禾郛鼓店态锭卡n-0.5^(n-1) =4*(0.5^1+0.5^2+0.5^3+....+0.5^n)-n*0.5^(n-1) 计算到这儿，你也明白了，我就不计算了
已知数列an=（2n+1）*2^n次方，求前n项和Sn
解：
Sn=3·2+5·2²+7·2³+...+(2n+1)·2ⁿ
2Sn=3·2²+5·2³+...+(2n-1)·2ⁿ+(2n+1)·2ⁿ⁺¹
Sn-2Sn=-Sn=3·2+2·2²+...+2·2ⁿ-(2n+1)·2ⁿ⁺¹
=2+2²+...+2ⁿ⁺¹-(2n+1)·2ⁿ⁺¹
=2·(2ⁿ⁺¹-1)/(2-1) -(2n+1)·2ⁿ⁺¹
=(1-2n)·2ⁿ⁺¹-2
Sn=(2n-1)·2ⁿ⁺¹+2
an=(2^n-1)*n，求{an}的前n项和
an=(2^n-1)×n=2^n×n-n
Sn=[1×2-1]+[2×2^2-2]+[3×2^3-3]+... ...+[(n-1)×2^(n-1)-(n-1)]+[n×2^n-n]
=[1×2+2×2^2+3×2^3+... ...+(n-1)×2^(n-1)+n×2^n]-[1+2+3+... ...(n-1)+n]
=[1×2+2×2^2+3×2^3+... ...+(n-1)×2^(n-1)+n×2^n]-1/2×n×(n+1)
设 Y=1×2+2×2^2+3×2^3+... ...+(n-1)×2^(n-1)+n×2^n
2Y= 1×2^2+2×2^3+3×2^4... ...+(n-2)×2^(n-1)+(n-1)×2^n+n×2^(n+1)
Y-2Y=2+2^2+2^3+2^4+... ...+2^(n-1)+2^n-n×2^(n+1)
-Y=2^n-1-n×2^(n+1)
Y=-[2^n-1-2n×2^n]
Y=(2n-1)×2^n+1
原式=Sn=(2n-1)×2^n+1-1/2×n×(n+1)