已知等差数列{an}中，a2=9，a5=21．（1）求{an}的通项公式；（

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问题解答题

已知等差数列{a_n}中，a₂=9，a₅=21．

（1）求{a_n}的通项公式；

（2）令b_n=2^a_n，求数列{b_n}的前n项和S_n．

答案

（1）设数列{a_n}的公差为d，由题意得

a+d=9

a1+4d

=21

解得a₁=5，d=4，

∴{a_n}的通项公式为a_n=4n+1．

（2）由a_n=4n+1得

b_n=2⁴ⁿ⁺¹，

∴{b_n}是首项为b₁=2⁵，公比q=2⁴的等比数列．

∴S_n=

25( 24n-1)

24-1

32×(24n-1)

．

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