1. The sum S of the first n positive in tegers can by the following formula:
S=n(n+1) / 2
(a)(i) Find the value of 1+2+3+...+40.
(ii)Find the value of 2+4+6+...+40.
(b) By using the results in (a), find the value of 1+3+5+...+39.
*please show the process, thanks
f.2 maths question
2007-03-03 6:24 am
回答 (3)
2007-03-03 6:31 am
✔ 最佳答案
ai)1+2+3+...+40=40(40+1)/2
=40*41/2
=820
aii)2+4+6+...+40
=2*(1+2+3+...+20)
=2*20(20+1)/2
=420
b)1+3+5+...+39
=(1+2+3+...+40)-(2+4+6+...+40)
=820-420
=400
2007-03-03 6:58 pm
1.
1+2+3+...+40
=(1+40)x40/2
=41x40/2
=820
2.
2+4+6+...+40
=(2+40)x((40-2)/2+1)/2
=42x20/2
=420
3.
1+3+5+...+39
=(1+2+3+...+40)-(2+4+6+...+40)
=820-420
=400
1+2+3+...+40
=(1+40)x40/2
=41x40/2
=820
2.
2+4+6+...+40
=(2+40)x((40-2)/2+1)/2
=42x20/2
=420
3.
1+3+5+...+39
=(1+2+3+...+40)-(2+4+6+...+40)
=820-420
=400
2007-03-03 6:32 am
(a)(i) Find the value of 1+2+3+...+40.
S=n(n+1) / 2
s=40(41)÷2
=820
(ii)Find the value of 2+4+6+...+40.
s=(首項+末項)項數 ÷ 2
=(2+40)20 ÷ 2
=420
(b) By using the results in (a), find the value of 1+3+5+...+39.
s=(首項+末項)項數 ÷ 2
=(1+39)20 ÷ 2
=400
S=n(n+1) / 2
s=40(41)÷2
=820
(ii)Find the value of 2+4+6+...+40.
s=(首項+末項)項數 ÷ 2
=(2+40)20 ÷ 2
=420
(b) By using the results in (a), find the value of 1+3+5+...+39.
s=(首項+末項)項數 ÷ 2
=(1+39)20 ÷ 2
=400
參考: king
收錄日期: 2021-04-12 23:23:31
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