数列Cn=(n乘以2的n次方)再除以（n的平方+3n+2）,它的前n项和是多少?

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变一下形.cn=[（n+1）*2^n-2^n]/[(n+1)*(n+2)]=2^n/(n+2)-2^n/[(n+1)*(n+2)]
1/(n+1)*(n+2)=1/(n+1)-1/(n+2)
cn=2^n/(n+2)-2^n*[1/(n+1)-1/(n+2)]=2*[2^n/(n+2)]-2^n/(n+1)=2^(n+1)/(n+2)-2^n/(n+1)
令an=2^n/(n+1)
很容易看出a(n+1)=2^(n+1)/(n+2)
所以cn=a（n+1）-an
cn的前n项和就等于a（n+1）-a1=2^(n+1)/(n+2)-1
注：^代表次方,2^n就是2的n次方./是除,*是乘以.
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供参考答案1:
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