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Math hard question

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What is 14+17+20+23+....+317

Guest Oct 19, 2017
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#1
+302
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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 = $$16881$$

Mathhemathh  Oct 20, 2017
#2
+1

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

Guest Oct 20, 2017
#3
+78557
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We have

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

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

( [ an - 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

CPhill  Oct 20, 2017

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