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當n=2,推演如下:
命f(m)=1*2+2*3+...+m(m+1)
故0.5f(m)=1*2/2+2*3/2+...+m(m+1)/2
明顯的, 第k項是第k個三角形數
因此,
f(m)/2=1+(1+2)+(1+2+3)+..+(1+2+3+..+m)
f(m)/2=1*m+ 2*(m-1) +3*(m-2) +..+m*1
f(m)/2=1m + 2m-1*2 + 3m-2*3 +..+ m
f(m)/2=m^2(m+1)/2 - 1*2 - 2*3 - 3*4 -..- (m-1)m -m(m+1)+m(m+1)
f(m)/2=m^2(m+1)/2 - f(m) + m(m+1)
3f(m)/2=m^2(m+1)/2 + m(m+1)
3f(m)=m^3+3m^2+2m
f(m)=m(m+1)(m+2)/3
Q.E.D.
至於n>2的情況, 就須依賴大家幫忙啦
1*2*3*4*...*n + 2*3*4*5*...*(n+1) + ..+ m(m+1)(m+2)..(m+n)後續
2007-07-30 12:07 am
回答 (2)
2007-08-08 12:32 pm
✔ 最佳答案
我認為以上數列應為1 × 2 × 3 × 4 × ... × n + 2 × 3 × 4 × 5 × ... × (n + 1) + ... + m(m + 1)(m + 2) ... (m + n – 1)
因為前2組有n項相乘，所以最後一組應有n項。
設a_r = r(r + 1)(r + 2) ... (r + n)，
所以a_r – a_(r – 1)
= r(r + 1)(r + 2) ... (r + n) – (r – 1)r(r + 1) ... (r – 1 + n)
= r(r + 1)(r + 2) ... (r – 1 + n)[(r + n) – (r – 1)]
= (n + 1)r(r + 1)(r + 2) ... (r + n – 1)
因此1 × 2 × 3 × 4 × ... × n + 2 × 3 × 4 × 5 × ... × (n + 1) + ... + m(m + 1)(m + 2) ... (m + n – 1)
=Σ(r = 1 to m)[r(r + 1)(r + 2) ... (r + n – 1)]
= (1/(n + 1))Σ(r = 1 to m)[a_r – a_(r – 1)]
= (1/(n + 1))(a_m – a_0)
= (a_m)/(n + 1)
= (m(m + 1)(m + 2) ... (m + n))/(n + 1)
參考: 修改自BREAKTHROUGH Pure Mathematics－ALGEBRA Chapter 7 Finite Sequences and Series P.244 Example 7.8(a)
2007-08-08 5:37 pm
(m(m + 1)(m + 2) ... (m + n))/(n + 1)
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