Math hard question

This is equal to $14+(14+3)+(14+3*2)...$

Separate the 14s: $14*102+3+3*2+...+3*101$

Factor: $14*102+3(1+2+...+101)$

Evaluate: $1428+3(1+2...+101)$

Sum of consecutive integers = ${n(n+1)\over2}$

Evaluate again: $1428+3({101(102)\over2})$

$1428+3*5151$

$1428+15453$

Final answer = $16881$ 

Number of terms = [317 - 14] / 3  +  1 = 102

[F + L] / 2  x 102 =, where F=First, L=Last.

[14 + 317] / 2  x 102 =16,881

We have 

[a₁ + a_n ] ( [ a_n - a₁ ] / 3 + 1 ) / 2

Where a₁ and a_n are the first and last terms to be summed  and 

( [ a_n - a₁ ] / 3 + 1 ) / 2     are the number of pairs of equal sums to be added together

So we have 

[ 14 + 317 ] ( [317 - 14 ] / 3  + 1 )  / 2  =  16881