Tn=3n+2為一等差數列的通項，求首12項之和

Tn=3n+2為一等差數列的通項，求首12項之和
[3n(n+1)/2]+2n
[3(12)(13)/2]+2(12)
=258

求1-300內所有13的倍數之和
[(13+299)x23]/2
=3588

T(1)=3x1+2=5
T(2)=3x2+2=8
T(3)=3x3+2=11
T(4)=3x4+2=14
T(5)=3x5+2=17
T(6)=3x6+2=20
T(7)=3x7+2=23
T(8)=3x8+2=26
T(9)=3x9+2=29
T(10)=3x10+2=32
T(11)=3x11+2=35
T(12)=3x12+2=38

Tn=3n+2
T(n+1)=3(n+1)+2
T1+T2+T3+...+T12=(3+2)+(6+2)+(9+2)+...+(36+2)
=2x12+3(1+2+3+...+12)
=24+3[12(12+1)/2]
=24+3(78)
=24+234
=258

The lagrest number that can divide by 13 and smaller than 300 is 299
13+26+39+...+299
=13(1+2+3+...+23)
=13(23x(23+1)/2)
=13(23x12)
=3588

2:3=2:a 2:3=3:b 2:3=4.5:c 2:3=6.75:d 2:3=10.125:e
a=3 b=4.5 c=6.75 d=10.125 e=15.1875
2+a+b+c+d+e=41.5625

(15-6)a=38-11 a is the common different
a=3
T1=11-3x(6-1)
=11-15
=-4
Tn=3n-7
T1+T2+...+T18=(3-7)+(6-7)+(9-7)+...+(54-7)
=3(1+2+3+...+18)-18x7
=3[18(18+1)/2]-126
=27x19-126
=513-126
=387