1+2+3+4+5+6+7+8+9.....+97+98+99+100(急..)

By

T(n) = a + (n - 1)d

100 = 1 + (n - 1)(1)

n = 100

So,by

S(n) = n/2 [ 2a + (n - 1)d ]

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50.......+100

= 100/2 [ 2(1) + (100 - 1)(1)]

= 50 x 101

= 5050

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50.......+100 = 5050

S(n) = n/2 [ 2a + (n - 1)d ]
如果可解釋此最好

1+2+3+4+5+6+7+8+9.....+97+98+99+100
1+100
= -----------*100
2
= 5050

設1+2+3+4+5+6+7+8+9.....+97+98+99+100=S,
則100+99+98+97+.....+9+8+7+6+5+4+2+1=S
......1+.2+..3+.....+98+99+100
+100+99+98+.....+..3+..2+...1
_________________________
..101+101+101+...+101+101(100個101)=2S
101X100=2S
101X50=S
S=5050
所以1+2+3+4+5+6+7+8+9.....+97+98+99+100=5050
PS:直式上的.是為了令直式更整齊,可以不理會

2007-04-08 20:35:57 補充：
Where does T(n) = a + (n - 1)d come from? I think you are just copying...

=(1+100)*100/2
=101*100/2
=10100/2
=5050

連續加法公式:(首項+尾項)*項數/2
=(1+100)*100/2
=101*50
=5050

[(頭項+尾項)x項數/2]
1+2+3+4+5+6+7+8+9.....+97+98+99+100

[(1+100)x100/2]
=5050

(1+100)+(2+99)+(3+98)+................+(50+51)
101+101+101+101+101+............+101.....(50)
101x50
=5050

1+2+3+4+5+6+7+8+9.....+97+98+99+100
=(1+100)x100/2 [(頭項+尾項)x項數/2]
=101x50
=5050

Ans: (1+100)x(100/2)=5050

1+2+3+4+5+6+7+8+9.....+97+98+99+100= 100 numbers
100/2= 50

For example: 1+2+3+4+5+6+7+8+9+10=?
1+2+3+4+5+6+7+8+9+10= 10 numbers
(1+10)x(10/2)=55

all are the same formula

(頭項+尾項) x 項數 / 2

=(1+100) x 100 / 2

=5050