1+4+7+...+ (3n - 2) = n/2 (3n - 1)

Answer 1

1+4+7+.. (3n-2) = (n/2)(3n-1)

it's an arithmetic series

each term is 3 more than the preceding term

an = a1 + (n-1)d where d is the common difference = 3

an = 1 +3(n-1)

= 1+3n-3

= 3n-2

for n terms the sum of the series

Sn= (n/2)(a1+an) = half the sum of 1st & last terms (Gauss' formula)

= (n/2)(1+3n-2)

= (n/2)(3n-1)

Sn = 1+4+7+...+(3n-2)

Sn+1 = 1+4+7+...(3n-2)+(3n+1)

Sn+1 =1+4+7+...+3n+1

= 1+4+7+...+(3(n+1)-2)

=1+4+7+...+3n+1

replace n with k

use mathematical induction to prove Gauss' formula

basic step

a1 = 1

s1 = 1 = (n/2)(3n-1) =(1/2)(3-1) = 2/2 = 1

a2 = 4

s2 =1+4=5 = (n/2)(3n-1) = (2/2)(6-1) = 6

assume Sk = (k/2)(3k-1)

try to show Sk+1 = ((k+1)/2)(3(k+1)-1)

Sk+1 = Sk+ak+1

= (k/2)(3k-`1) +3(k+1)-2

= (k/2)(3(k+1)-4) + 3(k+1)-2

= (k/2)(3(k+1)-4) + (3/2)(k+1)

= ((k+1)/2)(3(k+1)-(3/2)(+1) + (3/2)(k+1)

= ((k+1)/2)(3(k+1)-1)

QED