(√ 1)²+1=2,s1=√ 1/2;(√ 2)²+1=3,S2=√ 2/2;(√ 3)²+1=4,S3=√3/2……①请用含有n(n是正整数)的等式表示上述变化规律②推算出OA10的长③求出S1²+S2²+S3²+````+S10
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![(√ 1)²+1=2,s1=√ 1/2;(√ 2)²+1=3,S2=√ 2/2;(√ 3)²+1=4,S3=√3/2……①请用含有n(n是正整数)的等式表示上述变化规律②推算出OA10的长③求出S1²+S2²+S3²+````+S10](/uploads/image/z/2492332-52-2.jpg?t=%EF%BC%88%E2%88%9A+1%EF%BC%89%26%23178%3B%2B1%EF%BC%9D2%2Cs1%EF%BC%9D%E2%88%9A+1%2F2%3B%28%E2%88%9A+2%29%26%23178%3B%2B1%EF%BC%9D3%2CS2%EF%BC%9D%E2%88%9A+2%2F2%EF%BC%9B%EF%BC%88%E2%88%9A+3%EF%BC%89%26%23178%3B%2B1%EF%BC%9D4%2CS3%EF%BC%9D%E2%88%9A3%2F2%E2%80%A6%E2%80%A6%E2%91%A0%E8%AF%B7%E7%94%A8%E5%90%AB%E6%9C%89n%EF%BC%88n%E6%98%AF%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E7%9A%84%E7%AD%89%E5%BC%8F%E8%A1%A8%E7%A4%BA%E4%B8%8A%E8%BF%B0%E5%8F%98%E5%8C%96%E8%A7%84%E5%BE%8B%E2%91%A1%E6%8E%A8%E7%AE%97%E5%87%BAOA10%E7%9A%84%E9%95%BF%E2%91%A2%E6%B1%82%E5%87%BAS1%26%23178%3B%2BS2%26%23178%3B%2BS3%26%23178%3B%2B%60%60%60%60%2BS10%26%231)
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(√ 1)²+1=2,s1=√ 1/2;(√ 2)²+1=3,S2=√ 2/2;(√ 3)²+1=4,S3=√3/2……①请用含有n(n是正整数)的等式表示上述变化规律②推算出OA10的长③求出S1²+S2²+S3²+````+S10
(√ 1)²+1=2,s1=√ 1/2;(√ 2)²+1=3,S2=√ 2/2;(√ 3)²+1=4,S3=√3/2……
①请用含有n(n是正整数)的等式表示上述变化规律
②推算出OA10的长
③求出S1²+S2²+S3²+````+S10²的值.
(√ 1)²+1=2,s1=√ 1/2;(√ 2)²+1=3,S2=√ 2/2;(√ 3)²+1=4,S3=√3/2……①请用含有n(n是正整数)的等式表示上述变化规律②推算出OA10的长③求出S1²+S2²+S3²+````+S10
Sn=根号n/2,因为直角边长分别为1和根n,所以面积就是前面所写.
OA10=根10
S1²+S2²+S3²+````+S10²=(1+2+3+----10)/4=55/4