標題:
f.2 maths question
發問:
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
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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
其他解答:
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|||||(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 =40031C9A75CB3B14398
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