{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"omni-math","formal_name":"Omni-MATH","introduction":"Omni-MATH evaluates mathematical reasoning on Olympiad-level problems. Its official dataset contains 4,428 problems accompanied by domain and difficulty information.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"a105585d-dc90-5ac3-b3f6-292bb9620497","task_key":"test--a105585d-dc90-5ac3-b3f6-292bb9620497","task_revision_id":"1","upstream_id":"","short_description":"A sequence $(a_n)$ of real numbers is defined by $a_0=1$, $a_1=2015$ and for all…","config":"","split":"test","body":"{\"problem\":\"A sequence $(a_n)$ of real numbers is defined by $a_0=1$, $a_1=2015$ and for all $n\\\\geq1$, we have\\n$$a_{n+1}=\\\\frac{n-1}{n+1}a_n-\\\\frac{n-2}{n^2+n}a_{n-1}.$$\\nCalculate the value of $\\\\frac{a_1}{a_2}-\\\\frac{a_2}{a_3}+\\\\frac{a_3}{a_4}-\\\\frac{a_4}{a_5}+\\\\ldots+\\\\frac{a_{2013}}{a_{2014}}-\\\\frac{a_{2014}}{a_{2015}}$.\"}","display_format":"math","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/KbsdJames/Omni-MATH","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}