Research Space for Linear Algebra & Discrete Mathematics
n = 1: 1/(1*2) = 1/(1+1) hold假設n = k時命題成立即1/(1*2) + 1/(2*3) + ... + 1/(k*(k+1)) = k/(k+1)Consider n = k + 1:1/(1*2) + 1/(2*3) + ... + 1/(k*(k+1)) + 1/((k+1)(k+2))= k/(k+1) + 1/((k+1)(k+2))= [k(k+2)+1]/(k+1)(k+2)= (k^2 + 2k + 1)/(k+1)(k+2)= (k+1)^2/(k+1)(k+2)= (k+1)/(k+2)所以n = k+1亦成立
張貼留言
1 則留言:
n = 1: 1/(1*2) = 1/(1+1) hold
假設n = k時命題成立
即1/(1*2) + 1/(2*3) + ...
+ 1/(k*(k+1)) = k/(k+1)
Consider n = k + 1:
1/(1*2) + 1/(2*3) + ...
+ 1/(k*(k+1)) + 1/((k+1)(k+2))
= k/(k+1) + 1/((k+1)(k+2))
= [k(k+2)+1]/(k+1)(k+2)
= (k^2 + 2k + 1)/(k+1)(k+2)
= (k+1)^2/(k+1)(k+2)
= (k+1)/(k+2)
所以n = k+1亦成立
張貼留言