高一數學 ,請幫忙.

1.利用歸納法證明
1－1/2+1/3－1/4.........+1/(2n－1)－1/(2n)=1/(n+1)+1/(n+2)+....+1/(2n)
Sol
當n=1時
左=1－1/2=1/2
右=1*(1+1)=1/2
Son=1時為真
設n=k時為真即
1－1/2+1/3－1/4.........+1/(2k－1)－1/(2k)=1/(k+1)+1/(k+2)+....+1/(2k)
So
1－1/2+1/3－1/4.........+1/(2k－1)－1/(2k)+1/(2k+1)－1/(2k+2)
=1/(k+1)+1/(k+2)+....+1/(2k) +1/(2k+1)－1/(2k+2)
=2/(2k+2)+1/(k+2)+....+1/(2k) +1/(2k+1)－1/(2k+2)
=1/(k+2)+....+1/(2k) +1/(2k+1)+1/(2k+2)
Son=k+1時為真

2.設n為正整數，試証
1*n+2*(n－1)+3*(n－2)+......+(n－1)*2+n*1=n*(n+1)*(n+2)/6
Sol
當n=1時
左=1*1=1
右=1*(1+1)*(1+2)/6=1
Son=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
So
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+(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