摘要: 函数对任意都有成立. (Ⅰ)求和的值, (Ⅱ)数列满足条件,.试证:数列是等差数列.
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函数f(x)对任意x∈R都有f(x)+f(1-x)=
成立.
(Ⅰ)求和f(
)+f(
)(n∈N*)的值;
(Ⅱ)数列{an}满足条件;an=f(0)+f(
)+f(
)+…+f(
)+f(1),试证:数列{an}是等差数列.
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(Ⅰ)求和f(
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(Ⅱ)数列{an}满足条件;an=f(0)+f(
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