Sn是数列{an}的前n项和,an=(2n-1)i^n,求数列{an}的前2012项和
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Sn是数列{an}的前n项和,an=(2n-1)i^n,求数列{an}的前2012项和
![Sn是数列{an}的前n项和,an=(2n-1)i^n,求数列{an}的前2012项和](/uploads/image/z/7929703-55-3.jpg?t=Sn%E6%98%AF%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Can%3D%282n-1%29i%5En%2C%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8D2012%E9%A1%B9%E5%92%8C)
i⁴=(i²)²=(-1)²=1
i2012=(i⁴)^503=1^503=1
Sn=a1+a2+...+an=1×i+3×i²+5×i³+...+(2n-1)×iⁿ
iSn=1×i²+3×i³+...+(2n-3)×iⁿ+(2n-1)×i^(n+1)
Sn-iSn=(1-i)Sn=i+2i²+2i³+...+2iⁿ-(2n-1)×i^(n+1)
=2(i+i²+...+iⁿ) -i -(2n-1)×i^(n+1)
=2i×(iⁿ-1)/(i-1) -i -(2n-1)×i^(n+1)
Sn=2i(1-iⁿ)/(1-i)² -i -(2n-1)×i^(n+1)/(1-i)
=2i(1-iⁿ)/(1-2i-1) -i -(2n-1)×(1+i)×i^(n+1)/[(1-i)(1+i)]
=iⁿ-1 -i -(2n-1)(1+i)×i^(n+1)/2
令i=2012
S2012=i^2012 -i -(2n-1)(1+i)×i^2013 /2
=1 -i -(2n-1)(1+i)i/2
=1-i-(2n-1)(i-1)/2
=[2-2i-(2n-1)i+2n-1]/2
=(2n+1)/2 - (2n+1)i/2
i2012=(i⁴)^503=1^503=1
Sn=a1+a2+...+an=1×i+3×i²+5×i³+...+(2n-1)×iⁿ
iSn=1×i²+3×i³+...+(2n-3)×iⁿ+(2n-1)×i^(n+1)
Sn-iSn=(1-i)Sn=i+2i²+2i³+...+2iⁿ-(2n-1)×i^(n+1)
=2(i+i²+...+iⁿ) -i -(2n-1)×i^(n+1)
=2i×(iⁿ-1)/(i-1) -i -(2n-1)×i^(n+1)
Sn=2i(1-iⁿ)/(1-i)² -i -(2n-1)×i^(n+1)/(1-i)
=2i(1-iⁿ)/(1-2i-1) -i -(2n-1)×(1+i)×i^(n+1)/[(1-i)(1+i)]
=iⁿ-1 -i -(2n-1)(1+i)×i^(n+1)/2
令i=2012
S2012=i^2012 -i -(2n-1)(1+i)×i^2013 /2
=1 -i -(2n-1)(1+i)i/2
=1-i-(2n-1)(i-1)/2
=[2-2i-(2n-1)i+2n-1]/2
=(2n+1)/2 - (2n+1)i/2
Sn是数列{an}的前n项和,an=(2n-1)i^n,求数列{an}的前2012项和S2012
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