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In the following arrangement what would be the first number in 100th line?

 1   2   3  
 4   5   6   7  
 8   9  10  11  12  
13  14  15  16  17  18
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9
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Answer is

$(100 \times 103 \div 2) - 1 = 5149$

Because this sequence is

a(n) = binomial(n+1, 2) + n - 1 = n(n + 3)/2 - 1.

Since OP perhaps questions the correctness of this answer ... Try it online!

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    $\begingroup$ The question is about the first number in the 100th line, not the 100th number. $\endgroup$
    – Yousif
    Aug 20 '17 at 19:49
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    $\begingroup$ Well yes - that IS the first number of the 100th line. The 100th number is obviously just 100. $\endgroup$
    – Rubio
    Aug 20 '17 at 19:51
3
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Answer

5149

Explanation

Note that the third number in each row is a triangle number. Specifically, where n is the row number and the first row has n=1, the third number is (n+1)(n+2)/2. So in row 100, the third number is (101)(102)/2 = 5151, so the first number of row 100 is 5151-2 = 5149

Note also that my formula is mathematically equivalent to Rubio's:

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

However, my derivation of the formula, and therefore explanation of my derivation, is different from Rubio's.

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    $\begingroup$ Since your answer is the same as Rubio's, and your derivation of it is equivalent to Rubio's, why did you post it? $\endgroup$
    – Gareth McCaughan
    Aug 20 '17 at 21:58
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    $\begingroup$ @GarethMcCaughan Because my explanation, and the way i found the equation, was very different. It just happens that the math is isomorphic. $\endgroup$
    – benzene
    Aug 20 '17 at 22:01
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Answer:

1+(((101*102)/2))-3)=5149

Explanation:

First term is always 1.Then we calculate the difference between each lines first term is like 3,4,5,6.... So we calculate sum from 3 to n by using ((n+1*n+2)/2)-3 [3 is the sum of 1 and 2]. And then just add it with 1.It returns nth lines first term.

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    $\begingroup$ This is equivalent to another answer, posted a couple of hours before yours. Before posting an answer, please look and see whether it actually adds something beyond the other answers already posted. $\endgroup$
    – Gareth McCaughan
    Aug 20 '17 at 21:58
  • $\begingroup$ Sorry for the wrong explanation. I just write 100 in place of 102.and now my explanation same as @benzene.My thinking like @ benzene. $\endgroup$ Aug 22 '17 at 17:47

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