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修改两个题面

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weiye.wang 2024-01-30 21:14:39 +08:00
parent d9cb1f505d
commit a261324fc2
1 changed files with 6 additions and 4 deletions
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10
题库0.3/Problems.json
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@ -652697,7 +652697,7 @@
},
"024023": {
"id": "024023",
"content": "设 $a$、$b \\in R$, 若集合 $\\{1, a+b, a\\}=\\{0, b, \\dfrac{b}{a}\\}$, 则 $a^{2023}+b^{2023}=$\\blank{50}.",
"content": "设 $a$、$b \\in \\mathbf{R}$, 若集合 $\\{1, a+b, a\\}=\\{0, b, \\dfrac{b}{a}\\}$, 则 $a^{2023}+b^{2023}=$\\blank{50}.",
"objs": [],
"tags": [
"第一单元"
@ -652709,7 +652709,8 @@
"usages": [],
"origin": "自拟题目",
"edit": [
"20240125\t毛培菁"
"20240125\t毛培菁",
"20240130\t王伟叶"
],
"same": [],
"related": [],
@ -654426,7 +654427,7 @@
},
"024099": {
"id": "024099",
"content": "已知数列 $\\{a_n\\}$ 的首项 $a_1>0$, $a_{n+1}=\\dfrac{3 a_n}{2 a_n+1}$($n \\in N$, $n \\geq 1$), 且 $a_1=\\dfrac{2}{3}$.\\\\\n(1) 求证: $\\{\\dfrac{1}{a_n}-1\\}$ 是等比数列, 并求出 $\\{a_n\\}$ 的通项公式;\\\\\n(2) 求数列 $\\{\\dfrac{1}{a_n}\\}$ 的前 $n$ 项和 $T_n$.",
"content": "已知数列 $\\{a_n\\}$ 的首项 $a_1>0$, $a_{n+1}=\\dfrac{3 a_n}{2 a_n+1}$($n \\in \\mathbf{N}$, $n \\geq 1$), 且 $a_1=\\dfrac{2}{3}$.\\\\\n(1) 求证: $\\{\\dfrac{1}{a_n}-1\\}$ 是等比数列, 并求出 $\\{a_n\\}$ 的通项公式;\\\\\n(2) 求数列 $\\{\\dfrac{1}{a_n}\\}$ 的前 $n$ 项和 $T_n$.",
"objs": [],
"tags": [
"第四单元"
@ -654438,7 +654439,8 @@
"usages": [],
"origin": "自拟题目",
"edit": [
"20240125\t毛培菁"
"20240125\t毛培菁",
"20240130\t王伟叶"
],
"same": [],
"related": [],