等比数列{an}的通项公式an=2ˆn则数列{an+n}的前n项和为
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![等比数列{an}的通项公式an=2ˆn则数列{an+n}的前n项和为](/uploads/image/z/1972354-58-4.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan%3D2%26%23710%3Bn%E5%88%99%E6%95%B0%E5%88%97%7Ban%2Bn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA)
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等比数列{an}的通项公式an=2ˆn则数列{an+n}的前n项和为
等比数列{an}的通项公式an=2ˆn则数列{an+n}的前n项和为
等比数列{an}的通项公式an=2ˆn则数列{an+n}的前n项和为
因为数列{an}为等比数列,且an=2^n,所以an+n=2^n+n,所以an的前n项和Sn=a1(1-q^n)/(1-q)=2(1-2^n)/(1-2)=-2(1-2^n),所以数列{an+n}的前n项和Sn'=-2(1-2^n)+n(n+1)/2.