Question

是否存在常數(shù)a、b,使等式1×n+2(n-1)+3(n-2)+…+(n-2)×3+(n-1)×2+n×1=n(n+a)(n+b)對一切正整數(shù)N^*成立,證明你的結(jié)論.

Answer 1

分析：先取n=1、2,探求a、b的值,然后用數(shù)學(xué)歸納法證明對一切的n∈N^*,a、b所確立的等式都成立.

解：令n=1,得1=(1+a)(1+b),令n=2得4=(2+a)(2+b),

整理解得a=1,b=2.

下面利用數(shù)學(xué)歸納法證明等式1×n+2(n-1)+3(n-2)+…+(n-2)×3+(n-1)×2+n×1=n(n+1)(n+2).

證明：①當(dāng)n=1時,原等式成立.

②假設(shè)n=k時,1×k+2(k-1)+3(k-2)+…+(k-2)×3+(k-1)×2+k×1=k(k+1)(k+2)成立.

當(dāng)n=k+1時,1×(k+1)+2(k+1-1)+3(k+1-?2)+…+?(k+1-2)×3+(k+1-1)×2+(k+1)×1

=1×k+2(k-1)+3(k-2)+…+(k-2)×3+(k-1)×2+k×1+1+2+3+…+3+2+1=k(k+1)(k+2)+(k+1)(k+2)

=(k+1)(k+2)(k+)=(k+1)(k+2)(k+3),

即n=k+1時等式成立,由①②,知對任意n∈N^*,1×n+2(n-1)+3(n-2)+…+(n-2)×3+(n-1)×2+n×1

=n(n+1)(n+2).