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数列(an)各项的倒数组成一个等差数列,若a5=√2-1,a7=√2+1,求a3
人气:314 ℃ 时间:2020-07-26 13:14:23
解答
1/an = 1/a1 + (n-1)d
n=5
1/a5 = 1/a1 + 4d
1/(√2-1) = 1/a1 + 4d (1)
n=7
1/a7 = 1/a1 + 6d
1/(√2+1) = 1/a1 + 6d (2)
2(1) - (2)
a3=1/a1 + 2d
= 2/(√2-1) - 1/(√2+1)
= √2+3谢谢啦
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