a form 2 question about formula and substitution

(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

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 formula