✔ 最佳答案
1) When n = 1 ,
1 / (1*3*5) = 1(1+2) / (3(2*1+1)(2*1+3)) = 1/15 is true.Suppose it is true for n = k , i.e.
1/(1x3x5) + 1/(3x5x7) + ... + 1/((2k-1)(2k+1)(2k+3)) = k(k+2) / (3(2k+1)(2k+3))When n = k + 1 ,1/(1x3x5) + 1/(3x5x7) + ... + 1/((2k-1)(2k+1)(2k+3))
+ 1/((2(k+1)-1)(2(k+1)+1)(2(k+1)+3)) = k(k+2) / (3(2k+1)(2k+3))
+ 1/((2k+1)(2k+3)(2k+5))= ( k(k+2)(2k+5) + 3 ) / ( 3(2k+1)(2k+3)(2k+5) )
= ( k(k+2)(2k+1 + 4) + 3 ) / ( 3(2k+1)(2k+3)(2k+5) )
= ( (2k+1)k(k+2) + 4k(k+2) + 3 ) / ( 3(2k+1)(2k+3)(2k+5) )
= ( (2k+1)k(k+2) + 4k² + 8k + 3 ) / ( 3(2k+1)(2k+3)(2k+5) )
= ( (2k+1)k(k+2) + (2k+1)(2k+3) ) / ( 3(2k+1)(2k+3)(2k+5) )
= ( k(k+2) + (2k+3) ) / ( 3 (2k+3)(2k+5) )
= ( k² + 4k + 3) / ( 3 (2k+3)(2k+5) )
= (k+1)(k+3) / ( 3 (2k+3)(2k+5) ) is also true.By M.I. it is true for all positive integer.Hence
1/(21x23x25) + 1/(23x25x27) + 1/(25x27x29) + ... + 1/(67x69x71)= 1/(1x3x5) + 1/(3x5x7) + ... + 1/(67x69x71) ...... n = 34
- 1/(1x3x5) + 1/(3x5x7) + ... + 1/(19x21x23) ........n = 10 = 34(34+2) / (3(2*34+1)(2*34+3))
- 10(10+2) / (3(2*10+1)(2*10+3))= 16 / 34293
2)When n = 1 ,
2² = 2*1(1+1)(2*1+1)/3 = 4 is true.Suppose it is true for n = k , i.e.
2² + 4² + 6² + ... + (2k)² = 2k(k+1)(2k+1)/3When n = k + 1 ,2² + 4² + 6² + ... + (2k)² + (2(k+1))²
= 2k(k+1)(2k+1)/3 + (2(k+1))²
= (2k(k+1)(2k+1) + 12(k+1)² ) / 3
= (2(k+1) (k(2k+1) + 6(k+1) ) / 3
= (2(k+1) (2k² + 7k + 6) / 3
= 2(k+1) (k+2) (2k+3) / 3 is also true.By M.I. it is true for all positive integer.Hencei)
1² + 2² + 3² + ... + n²
= ( 2² + 4² + 6² + ... + (2n)² ) / 2²
= ( 2n(n+1)(2n+1)/3 ) / 2²
= n (n+1) (2n+1) / 6ii)
1² + 2² + 3² + ... + (2n)²
= ( 2² + 4² + 6² + ... + (4n)² ) / 2²
= ( 4n(2n+1)(4n+1)/3 ) / 2²
= n (2n+1) (4n+1) / 3iii)
(n+1)² + (n+2)² + ... + (2n)²
= 1² + 2² + 3² + ... + (2n)² - (1² + 2² + 3² + ... + n²)
= n (2n+1) (4n+1) / 3 - n (n+1) (2n+1) / 6
= ( 2n (2n+1) (4n+1) - n (n+1) (2n+1) ) / 6
= n(2n+1) ( 2(4n+1) - (n+1) ) / 6
= n (2n+1) (7n+1) / 6
2012-01-10 23:46:57 補充：
1² + 2² + 3² + ... + (2n)² - (1² + 2² + 3² + ... + n²) = (n+1)² + (n+2)² + ... + (2n)²
因為
1² + 2² + 3² + ... + (2n)² = (1² + 2² + 3² + ... + n²) + (n+1)² + (n+2)² + ... + (2n)²
就是這樣
