已知:(x+1)n=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+…+an(x-1)n(n≥2,n∈N*)(1)当n=5时,求a0+a
已知:(x+1)n=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+…+an(x-1)n(n≥2,n∈N*)(1)当n=5时,求a0+a1+a2+a3+a4+a...
已知:(x+1)n=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+…+an(x-1)n(n≥2,n∈N*)(1)当n=5时,求a0+a1+a2+a3+a4+a5的值.(2)设bn=a22n-3,Tn=b2+b3+b4+…+bn.试用数学归纳法证明:当n≥2时,Tn=n(n+1)(n-1)3.
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(1)当n=5时,
原等式变为(x+1)
5=a
0+a
1(x-1)+a
2(x-1)
2+a
3(x-1)
3+a
4(x-1)
4+a
5(x-1)
5令x=2得a
0+a
1+a
2+a
3+a
4+a
5=3
5=243
(2)因为(x+1)
n=[2+(x-1)]
n所以a
2=C
n2?2
n-2
bn==2=n(n-1)(n≥2)①当n=2时.左边=T
2=b
2=2,右边=
=2左边=右边,等式成立.
②假设当n=k(k≥2,k∈N
*)时,等式成立,即
Tk=那么,当n=k+1时,
左边=
Tk+bk+!=+(k+1)[(k+1)-1]=+k(k+1)=
k(k+1)(+1)===右边.
故当n=k+1时,等式成立.
综上①②,当n≥2时,
Tn=
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