请问怎么算的 f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1 ?1. 二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1 求f(X)解析:∵二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1f(x+1)=f(x)+2xf(1)=f(0)=1f(2)=f(1)+2?1=3f(3)=f(2)+2?2=7f(4)=f(3)+2?3=13
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![请问怎么算的 f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1 ?1. 二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1 求f(X)解析:∵二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1f(x+1)=f(x)+2xf(1)=f(0)=1f(2)=f(1)+2?1=3f(3)=f(2)+2?2=7f(4)=f(3)+2?3=13](/uploads/image/z/5252633-17-3.jpg?t=%E8%AF%B7%E9%97%AE%E6%80%8E%E4%B9%88%E7%AE%97%E7%9A%84+f%28n%29%3D1%2B2%281%2B2%2B3%2B%E2%80%A6%2Bn-1%29%3Dn%28n-1%29%2B1+%3F1.+%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3++f%28x%2B1%29-f%28x%29%3D2x++%E4%B8%94++f%280%29%3D1+%E6%B1%82f%28X%29%E8%A7%A3%E6%9E%90%EF%BC%9A%E2%88%B5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3++f%28x%2B1%29-f%28x%29%3D2x++%E4%B8%94++f%280%29%3D1f%28x%2B1%29%3Df%28x%29%2B2xf%281%29%3Df%280%29%3D1f%282%29%3Df%281%29%2B2%3F1%3D3f%283%29%3Df%282%29%2B2%3F2%3D7f%284%29%3Df%283%29%2B2%3F3%3D13)
请问怎么算的 f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1 ?1. 二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1 求f(X)解析:∵二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1f(x+1)=f(x)+2xf(1)=f(0)=1f(2)=f(1)+2?1=3f(3)=f(2)+2?2=7f(4)=f(3)+2?3=13
请问怎么算的 f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1 ?
1. 二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1 求f(X)
解析:∵二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1
f(x+1)=f(x)+2x
f(1)=f(0)=1
f(2)=f(1)+2?1=3
f(3)=f(2)+2?2=7
f(4)=f(3)+2?3=13
……
f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1
∴F(x)=x^2-x+1
请问怎么算的 f(n)=1+2(1+2+3+…+n-1)=n(n-1)+1 ?1. 二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1 求f(X)解析:∵二次函数f(x)满足 f(x+1)-f(x)=2x 且 f(0)=1f(x+1)=f(x)+2xf(1)=f(0)=1f(2)=f(1)+2?1=3f(3)=f(2)+2?2=7f(4)=f(3)+2?3=13
1+2+3+…+n-1=n(n-1)/2(等差数列求和)
1+2(1+2+3+…+n-1)=1+2*[n(n-1)/2]=n(n-1)+1
等差数列求和公式。