已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等
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![已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等](/uploads/image/z/2718699-51-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Ca2%3D-13%2Can%2B2-2an%2B1%2Ban%3D2n-6b1%3Da2-a1%3D-13-1%3D-14bn%3Da%28n%2B1%29-ana%28n%2B2%29-2a%28n%2B1%29%2Ban%3D2n-6a%28n%2B2%29-a%28n%2B1%29-a%28n%2B1%29%2Ban%3D2n-6a%28n%2B2%29-a%28n%2B1%29-%5Ba%28n%2B1%29%2Ban%5D%3D2n-6b%28n%2B1%29-bn%3D2n-6%E6%89%80%E4%BB%A5bn-b%28n-1%29%3D2%28n-1%29-6bn-b%28n-1%29%3D2%28n-1%29-6.b3-b2%3D2%2A2-6b2-b1%3D2%2A1-6%E4%BB%A5%E4%B8%8A%E7%AD%89)
已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等
已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6
b1=a2-a1
=-13-1
=-14
bn=a(n+1)-an
a(n+2)-2a(n+1)+an=2n-6
a(n+2)-a(n+1)-a(n+1)+an=2n-6
a(n+2)-a(n+1)-[a(n+1)+an]=2n-6
b(n+1)-bn=2n-6
所以bn-b(n-1)=2(n-1)-6
bn-b(n-1)=2(n-1)-6
.
b3-b2=2*2-6
b2-b1=2*1-6
以上等式相加得
bn-b1=2*1-6+2*2-6+.+2(n-1)-6
bn-b1=2*(1+2+3+.+n-1)-6(n-1)
bn-b1=n(n-1)-6n+6
bn-(-14)=n^2-n-6n+6
bn+14=n^2-7n+6
bn=n^2-7n-8
bn-b1=2*1-6+2*2-6+.+2(n-1)-6
bn-b1=2*(1+2+3+.+n-1)-6(n-1)
bn-b1=n(n-1)-6n+6
2为什么约掉了.
已知数列{an}满足a1=1,a2=-13,a(n+2)-2(an+1)+an=2n-6 1)设bn=A(n+1)-An,求数列{bn}的通项公式
已知数列{an}满足a1=1,a2=-13,an+2-2an+1+an=2n-6b1=a2-a1=-13-1=-14bn=a(n+1)-ana(n+2)-2a(n+1)+an=2n-6a(n+2)-a(n+1)-a(n+1)+an=2n-6a(n+2)-a(n+1)-[a(n+1)+an]=2n-6b(n+1)-bn=2n-6所以bn-b(n-1)=2(n-1)-6bn-b(n-1)=2(n-1)-6.b3-b2=2*2-6b2-b1=2*1-6以上等
1+2+3+.+n-1=(1+n-1)(n-1)/2 等差数列求和哦~所以 跟外面的2约了!
因为(1+2+3+......+n-1)=(n-1)(n-1+1)/2=n(n-1)/2