数列解答题 数列解答题
\u4eba\u6559\u724850\u9053\u6570\u5217\u89e3\u7b54\u9898\u53bb\u767e\u5ea6\u4e00\u4e0b\uff0c\u770b\u4e00\u4e0b\u603b\u7ed3\u4e00\u4e0b\uff0c\u65b9\u6cd5\u5f88\u91cd\u8981
\u7b2c n \u884c\u6709 2^(n-1) \u4e2a\u6570\uff0c
\u56e0\u6b64 an = 1+2+2^2+...+2^(n-1) = 2^n - 1 \u3002
\u8fd9 an \u4e2a\u6570\u7ec4\u6210\u9996\u9879\u4e3a 1\uff0c\u516c\u5dee\u4e3a 1\uff0c\u7684\u7b49\u5dee\u6570\u5217\uff0c
\u56e0\u6b64 Sn = 1/2*an*(an+1) = \uff082^n - 1\uff09*2^(n-1) \u3002
由题，数列｛abn}是公比为q的等比数列
得 a10平方=a1 * a46
又 数列｛an}是公差d不为0的等差数列
所以 (a1+9d)^2 = a1*(a1+45d)
求得 a1 = 3d
所以 q = a10/a1 = (a1+9d)/a1 = 4a1/a1 = 4
而 abn = a1+(bn-1)d = d*(bn+2)(利用等差)
且 abn = a1*4^(n-1) = 3d*4^(n-1)(利用等比)
所以 bn = 3*4^(n-1)-2
a是什么?
看不懂
老兄，你这题目真的没说清。a不清楚
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