由题设可得. 是怎么推出来的
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由题设可得. 是怎么推出来的
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![由题设可得. 是怎么推出来的](/uploads/image/z/7830479-47-9.jpg?t=%E7%94%B1%E9%A2%98%E8%AE%BE%E5%8F%AF%E5%BE%97.%26nbsp%3B+%26nbsp%3B%E6%98%AF%E6%80%8E%E4%B9%88%E6%8E%A8%E5%87%BA%E6%9D%A5%E7%9A%84)
(1)
a1=2
a2+a4=8
f(x) =[an-a(n+1)+a(n+2)]x+a(n+1)cosx-a(n+2)sinx
f'(x) =[an-a(n+1)+a(n+2)]-a(n+1)sinx-a(n+2)cosx
f'(π/2)=0
[an-a(n+1)+a(n+2)]-a(n+1) =0
a(n+2)-a(n+1) = a(n+1) -an
{a(n+1) -an}是等差数列, d=0
a(n+1) -an =a2-a1 (1)
a(n+1) -an =a3-a2 (2)
a(n+1) -an =a4-a3 (3)
(3)+(2)+2(1)
4[a(n+1) -an] = a4+a2-2a1
=8-4
a(n+1)-an = 1
an-a1 =n-1
an = n+1
a1=2
a2+a4=8
f(x) =[an-a(n+1)+a(n+2)]x+a(n+1)cosx-a(n+2)sinx
f'(x) =[an-a(n+1)+a(n+2)]-a(n+1)sinx-a(n+2)cosx
f'(π/2)=0
[an-a(n+1)+a(n+2)]-a(n+1) =0
a(n+2)-a(n+1) = a(n+1) -an
{a(n+1) -an}是等差数列, d=0
a(n+1) -an =a2-a1 (1)
a(n+1) -an =a3-a2 (2)
a(n+1) -an =a4-a3 (3)
(3)+(2)+2(1)
4[a(n+1) -an] = a4+a2-2a1
=8-4
a(n+1)-an = 1
an-a1 =n-1
an = n+1