已知an 求Sn(数列求和问题,要求用错位相减法)
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已知an 求Sn(数列求和问题,要求用错位相减法)
1.an=n×2ˆn 求Sn(错位相减法)
2.an=n×2ˆ(n-1)求Sn(错位相减)
3.an=n×(1/2)ˆn 求Sn(错位相减)
4.an=(2n-1)×2ˆn 求Sn(错位相减)
1.an=n×2ˆn 求Sn(错位相减法)
2.an=n×2ˆ(n-1)求Sn(错位相减)
3.an=n×(1/2)ˆn 求Sn(错位相减)
4.an=(2n-1)×2ˆn 求Sn(错位相减)
![已知an 求Sn(数列求和问题,要求用错位相减法)](/uploads/image/z/3984769-1-9.jpg?t=%E5%B7%B2%E7%9F%A5an+%E6%B1%82Sn%EF%BC%88%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E9%97%AE%E9%A2%98%2C%E8%A6%81%E6%B1%82%E7%94%A8%E9%94%99%E4%BD%8D%E7%9B%B8%E5%87%8F%E6%B3%95%EF%BC%89)
1.Sn=1×2+2×2^2+3×2^3+.+n×2^n (1)
2Sn= 1×2^2+2×2^3+.+(n-1)×2^n+n×2^(n+1) (2)
(2)-(1):Sn=-2-2^2-2^3-.-2^n+n×2^(n+1)
=-2(1-2^n)/(1-2)+n×2^(n+1)
=(n-1)×2^(n+1)+2
2.Sn=1+2×2+3×2^2+.+n×2^(n-1) (3)
2Sn= 1×2+2×2^2+...+(n-1)×2^(n-1)+n×2^n (4)
(4)-(3):Sn=-1-2-2^2-.-2^(n-1)+n×2^n
=-(1-2^n)/(1-2)+n×2^n
=(n-1)×2^n+1
3.Sn=1×(1/2)+2×(1/2)^2+3×(1/2)^3+.+n×(1/2)^n (5)
1/2Sn= 1×(1/2)^2+2×(1/2)^3+.+(n-1)×(1/2)^n+n×(1/2)^(n+1) (6)
(5)-(6):1/2Sn=1/2+(1/2)^2+(1/2)^3+.+(1/2)^n+n×(1/2)^(n+1)
=1+(n-2)/2×(1/2)^n
Sn=(n-2)×(1/2)^n+2
4.Sn=1×2+3×2^2+5×2^3+.+(2n-1)×2^n (7)
2Sn= 1×2^2+3×2^3+.+(2n-3)×2^n+(2n-1)×2^(n+1) (8)
(8)-(7):Sn=-2-2×2^2-2×2^3-.-2×2^n+(2n-1)×2^(n+1)
=-2-8[1-2^(n-1)]/(1-2)+(2n-1)×2^(n+1)
=(2n-3)×2^(n+1)+6
2Sn= 1×2^2+2×2^3+.+(n-1)×2^n+n×2^(n+1) (2)
(2)-(1):Sn=-2-2^2-2^3-.-2^n+n×2^(n+1)
=-2(1-2^n)/(1-2)+n×2^(n+1)
=(n-1)×2^(n+1)+2
2.Sn=1+2×2+3×2^2+.+n×2^(n-1) (3)
2Sn= 1×2+2×2^2+...+(n-1)×2^(n-1)+n×2^n (4)
(4)-(3):Sn=-1-2-2^2-.-2^(n-1)+n×2^n
=-(1-2^n)/(1-2)+n×2^n
=(n-1)×2^n+1
3.Sn=1×(1/2)+2×(1/2)^2+3×(1/2)^3+.+n×(1/2)^n (5)
1/2Sn= 1×(1/2)^2+2×(1/2)^3+.+(n-1)×(1/2)^n+n×(1/2)^(n+1) (6)
(5)-(6):1/2Sn=1/2+(1/2)^2+(1/2)^3+.+(1/2)^n+n×(1/2)^(n+1)
=1+(n-2)/2×(1/2)^n
Sn=(n-2)×(1/2)^n+2
4.Sn=1×2+3×2^2+5×2^3+.+(2n-1)×2^n (7)
2Sn= 1×2^2+3×2^3+.+(2n-3)×2^n+(2n-1)×2^(n+1) (8)
(8)-(7):Sn=-2-2×2^2-2×2^3-.-2×2^n+(2n-1)×2^(n+1)
=-2-8[1-2^(n-1)]/(1-2)+(2n-1)×2^(n+1)
=(2n-3)×2^(n+1)+6