a-maths!!!!!!!

(a)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/CrazyMI3.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Sep07/CrazyMI4.jpg

(a)試利用數學歸納法證明:對所有正整數n,
1²+2²+3²+...+n²= (1/6) n(n+1)(2n+1)
(1/6) (1)(1+1)(2+1) = 1 = 1²
assume 1²+2²+3²+...+k² = (1/6) k(k+1)(2k+1)
1²+2²+3²+...+k² + (k+1)²
= (1/6) k(k+1)(2k+1) + (k+1)²
= (1/6)(k+1)(2k^2 + k + 6k + 6)
= (1/6)(k+1)(2k^2 + 7k + 6)
= (1/6)(k+1)(2k + 3)(k + 2)
= (1/6) (k+1)(k+1+1) (2(k+1)+1)
by M.I., 1²+2²+3²+...+n²= (1/6) n(n+1)(2n+1)

(b)由此利用(a)結果,求下例各式之和:
(i) 1²+2²+3²+...+(2m)²
= (1/6)(2m)(2m+1)(4m+1)
= (m)(2m+1)(4m+1) / 3

(ii) 2²+4²+6²+...+(2m)²
= 2² 1² + 2² 2² + 2² 3² + ...+ 2² m²
= 2²(1² + 2² + 3² + ...+ m²)
= 4 (1/6) m(m+1)(2m+1)
= 2m(m+1)(2m+1)/3

(iii)1²+3²+5²+...+(2m-1)²
= 1²+2²+3²+...+(2m)² - [2²+4²+6²+...+(2m)²]
= (m)(2m+1)(4m+1)/3 - 2m(m+1)(2m+1)/3
= (m)(2m+1)(2m-1)/3

2007-09-05 22:03:02 補充：
慢左, 不過...請多多支持