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