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a form 2 question about formula and substitution

發問:

The sum S of the first n positive integers can be calculated 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. I need step for each answer! 更新: p.s. (ii) Find the value of 2+4+6+...+40

最佳解答:

(a)(i) S=n(n+1)/2 Put n=40, 1 + 2 + 3 + ... ... + 40 =40(40+1)/2 =20×41 =820 (ii) 2 + 4 + 6 + ... ... + 40 =2 (1 + 2 + 3 + ... ... + 20) =2×20(20+1)/2 [Put n=20] =20×21 =420 (b) 1 + 3 +5 + ... ... + 39 =(1 + 2 + 3 + ... ... + 40) - (2 + 4 + 6 + ... ... + 40) =840 - 420 =420 2007-11-22 20:58:37 補充： (b)1＋3＋5＋... ...＋39=(1＋2＋3＋... ...＋40) - (2＋4＋6＋... ...＋40)=820 - 420=400

其他解答:

it had gave you the formula, so use the formula (a)(i)Find the value of 1+2+3+...+40. S=n(n+1)/2 S=40(40+1)/2 S=40X41/2 S=20X41 S=820 (ii)Find the value of 2+4+6+...+40. S=n(n+1)/2 S=40(40+1)/2X2 S=40X41/4 S=10X41 S=410 (b) By using the result in (a), find the value of 1+3+5+...+39 820-(1+2+3+...20) =820-[20(20+1)/2} =820-(20X21/2) =820-210 =610 All these are triangle formula31C9A75CB3B14398