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数列an=n^2求和
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数列an=n^2 求和
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答案和解析
an = n²
Sn = 1² + 2² + 3² + .+ n² = n(n+1)(2n+1)/6
归纳法证明:
n = 1,1×(1+1)×(2×1+1)/6 = 6/6 = 1,求和公式正确
设 n = k 时,Sk = 1² + 2² + 3² + .+ k² = k(k+1)(2k+1)/6 成立.
S(k+1) = k(k+1)(2k+1)/6+(k+1)²
= (k+1)[k(2k+1)/6+(k+1)]
= (k+1)[k(2k+1)+6k+6]/6
= (k+1)[2k²+7k+6]/6
= (k+1)[(k+2)(2k+3]/6
= (k+1)[(k+1)+1][2(k+1)+1]/6
得证.
Sn = 1² + 2² + 3² + .+ n² = n(n+1)(2n+1)/6
归纳法证明:
n = 1,1×(1+1)×(2×1+1)/6 = 6/6 = 1,求和公式正确
设 n = k 时,Sk = 1² + 2² + 3² + .+ k² = k(k+1)(2k+1)/6 成立.
S(k+1) = k(k+1)(2k+1)/6+(k+1)²
= (k+1)[k(2k+1)/6+(k+1)]
= (k+1)[k(2k+1)+6k+6]/6
= (k+1)[2k²+7k+6]/6
= (k+1)[(k+2)(2k+3]/6
= (k+1)[(k+1)+1][2(k+1)+1]/6
得证.
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