Question

已知数列an满足a1=1,an-2an-1-2^(n-1)=0,求它的前n项和Sn

Answer 1

an-2a(n-1)=2^(n-1)
an/2^n-a(n-1)/2^(n-1)=1/2
所以{an/2^n}是公差为1/2的等差数列
an/2^n=a1/2^1+(n-1)*(1/2)=/12+n/2-1/2=n/2
an=n*2^(n-1)
Sn=1+2*2+3*2^2+.....+n*2^(n-1)
2Sn=2+2*2^2+3*2^3+....+n*2^n
Sn-2Sn=1+2+2^2+.....+2^(n-1)-n*2^n
-Sn=(2^n-1)/(2-1)-n*2^n
Sn=n*2^n-2^n+1
=(n-1)*2^n+1

Answer 2

解：an-2a(n-1)=2^n-1
∴ an=2a(n-1)+2^n-1
两边同时加上2^n-1
则 an+2^n-1=2[a(n-1)+2^n-1]
设an+2^n-1=bn
b1=a1+2^0=2
bn=2b(n-1)
推出bn=2^n
则 an=bn-2^n-1=2^n-2^n-1
∴sn=a1+a2+a3+......an
=2^1-2^0 + 2^2-2^1 +.......+ 2^n-2^n-1
=2^n-2^0=2^n - 1
求sn时用倒序相加法亦可