help!~~~~~ amaths!!!!

2007-09-15 3:54 am
請列明步驟 ^^ thx ~~~

1 Prove by M.I. that

1 / ( 1x2x3) + 1/(2x3x4) + 1/(3x4x5)+......+1/ n(n+1)(n+2) = n(n+3) / 4(n+1)(n+2)

for all positive integerss n .


2 . Prove by M. I . that

1^2 + 2^2 + 3^2 + ......+ n^2 = 1/ 6 ( n ) ( n+1)(2n+1)

for all positive integers n.

Hence, find the sum of
( i ) 3x5 +7x9 + 11x13 +.......+ 95x97
更新1:

question number 2 , I have already proved that 1^2 + 2^2 + 3^2 + ......+ n^2 = 1/ 6 ( n ) ( n+1)(2n+1) is ture for all positive integers n. However , i don't know how to find 3x5 +7x9 + 11x13 +.......+ 95x97 = ????

回答 (2)

2007-09-15 5:17 am
我只識中文
1.設p(n)為命題,
p(n)=1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+......+1/n(n+1)(n+2)= n(n+3)/4(n+1)(n+2)
當n=1
左p(1)=1/(1x2x3)=1/6
右p(1)=(1x4)/4(1x2x3)=1/6
左p(1)=右p(1)
p(1)成立
假設p(k)成立,1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+......+1/k(k+1)(k+2)=k(k+3)/4(k+1)(k+2)
當n=k+1
p(k+1)=1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+......+1/k(k+1)(k+2)+1/(k+1)[(k+1)+1][(k+1)+1]
=k(k+3)/4(k+1)(k+2)+1/(k+1)(k+2)(k+3)
=[1/4(k+2)(k+3)]*[k(k+3)^2/(k+1)+4/(k+1)]
=[1/4(k+2)(k+3)]*[(k^3+6k^2+9k+4)/(k+1)]
=[1/4(k+2)(k+3)]*[(k+1)(k+1)(k+4)/(k+1)]
=[1/4(k+2)(k+3)](k+1)(k+4)
=(k+1)[(k+1)+3]/{4[(k+1)+1][(k+1)+2]}
=右p(k+1)
p(k+1)成立

根據............,命題p(n)成立
2b 唔識計


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