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#220f4a9c-57a6-4ab7-8694-86f9e8a3b787中等解答题等比数列的概念与通项数列

26. 已知等比数列{an}\{a_{n}\}的前nn项和为SnS_{n}a1=1a_{1}=-1S6S3=2627\frac{S_6}{S_3}=\frac{26}{27}
(1)求等比数列{an}\{a_{n}\}的通项公式;
(2)求a12+a22++an2{a}_{1}^{2}+{a}_{2}^{2}+\ldots +{a}_{n}^{2}的值.