#### soccerisgreat

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- Thread starter soccerisgreat
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\(\displaystyle \L\\\sum_{k=1}^{10}k=55\)

You can use \(\displaystyle \L\\\frac{n(n+1)}{2}=1000\)

Solve for n. That should give you a good look at what row it's in.

Round your answer up.

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1, 3, 6, 10, 15, ...

the nth term of this sequence is n(n+1)/2

to find the row where 1000 resides ...

set n(n+1)/2 = 1000

n(n+1) = 2000

n should be close to the square root of 2000 ... approx 44.7

check the n-values close to 44.7 using the nth term above ...

44*45/2 = 990

45*46/2 = 1035

1000 lies in the 45th row.

The first numbers in the triangular rows are 1, 2, 4, 7, 11,...etc. and are defined by N = (n^2 - n + 2)/2soccerisgreat said:

Setting this expression equal to 1000 yields an n = 45.2 meaning that the number 1000 is in the 45th row.