Prove by mathematical induction that the following propositions are true for all positive integers n
1 x n + 2 x (n-1) + 3 x (n-2) + ... + (n-1) x 2 + n x 1 = n(n+1)(n+2)/6
數學歸納法問題求解?
2018-05-06 12:52 pm
回答 (2)
2018-05-06 4:12 pm
1*n+2*(n－1)+3*(n－2)+... +(n－1)*2+n*1=n(n+1)(n+2)/6
Sol
當n=1時
1*1=1
1*(1+1)*(1+2)/6=1*2*3/6=1
So n=1時為真
設n=k時為真
即1*k+2*(k－1)+3*(k－2)+... +(k－1)*2+k*1=k(k+1)(k+2)/6
1*(k+1)+2*k+3*(k－1)+... +(k－1)*3+k*2+(k+1)*1
=[1*k+2*(k－1)+3*(k－2)+... +(k－1)*2+k*1]+[1*1+2*1+3*1+…+(k+1)*1]
=k(k+1)(k+2)/6+(k+1)*(k+2)/2
=(k+1)(k+2)/6*(k+3)
=(k+1)(k+2)(k+3)/6
So n=k+1時為真
Sol
當n=1時
1*1=1
1*(1+1)*(1+2)/6=1*2*3/6=1
So n=1時為真
設n=k時為真
即1*k+2*(k－1)+3*(k－2)+... +(k－1)*2+k*1=k(k+1)(k+2)/6
1*(k+1)+2*k+3*(k－1)+... +(k－1)*3+k*2+(k+1)*1
=[1*k+2*(k－1)+3*(k－2)+... +(k－1)*2+k*1]+[1*1+2*1+3*1+…+(k+1)*1]
=k(k+1)(k+2)/6+(k+1)*(k+2)/2
=(k+1)(k+2)/6*(k+3)
=(k+1)(k+2)(k+3)/6
So n=k+1時為真
收錄日期: 2021-04-23 20:58:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180506045221AADwuQU