在等差数列{an},a3+a8+a14+a9=64,求前16项之和S16(要过程)
人气:469 ℃ 时间:2020-05-27 09:52:24
解答
a1+a16=a2+a15=...=a8+a9=[(a3+a14)+(a8+a9)]/2=32
S16=a1+a2+...+a16=8(a8+a9)=256
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