Find the sum of the arithmetic series 4+7+10+13+...+(3n+1)

Answer 1

Sn = (n/2)(a1+an) = (n/2)(4+3n+1) = (n/2)(3n+5) = 3n^2/2 + 5n/2

sum of a1 through an = (3/2)n^2 + 5n/2

S3 = (3/2)9 + 5(3)/2 = 27/2 + 15/2 = 42/2 = 21

a1=4

a2=7

a3 =10

s3 = 4+7+10 = 21

Answer 2

There are four sum formulas you need:

(where c is constant)

∑ⁿ_i=1(a_i + b_i) = ∑ⁿ_i=1 (a_i) + ∑ⁿ_i=1 (b_i)

∑ⁿ_i=1 c = cn

∑ⁿ_i=1 (ca_i) = c ∑ⁿ_i=1(a_i)

∑ⁿ_i=1 (i ) = n(n+1)/2

Now to solve the problem ∑ⁿ_i=1(3i + 1) = 4 + 7 + 10 + ...+ (3n + 1) using the formula above:

∑ⁿ_i=1(3i + 1) = ∑ⁿ_i=1 (3i) + ∑ⁿ_i=1 1

= 3•∑ⁿ_i=1i + (1)(n)

= 3•n(n+1)/2 + n

= (3n²+ 3n + 2n) / 2

= (3n²+ 5n) /2