1 2 3 4 .... 1002 what is the sum of the sequence

1 answer:

3 0

As far as i know it is related to Gauss.
Write the sequences forward and backward first.

1 +2 +3 +.....+1002
1002+1001+1000+.....+1
--------------------------------------... Adding them
1003+1003+......(1002 times)
=1002x1003
But this contains the series twice.
So, the sum is = 1002x1003/2=501x1003=502503. answer

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Consider M, N, and P. collinear points on MP.

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Answer:

See explanation

Step-by-step explanation:

There are three possible cases:

1. Point N lies between M and P, then MN + NP = MP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{NP}-\dfrac{MN}{MN+NP}=\dfrac{MN(MN+NP)-MN\cdot NP}{NP(MN+NP)}=\\ \\=\dfrac{MN^2+MN\cdot NP-MN\cdot NP}{NP(MN+NP)}=\dfrac{MN^2}{NP(MN+NP)}$

2. Point N lies to the right from point P, then MP + PN = MN. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MP+PN}{NP}-\dfrac{MP+PN}{MP}=\dfrac{MP}{NP}+1-1-\dfrac{NP}{MP}=\dfrac{MP^2-NP^2}{NP\cdot MP}$

3. Point N lies to the left from point M, then NM + MP = NP. Consider needed difference:

$\dfrac{MN}{NP}-\dfrac{MN}{MP}=\dfrac{MN}{MN+MP}-\dfrac{MN}{MP}=\dfrac{MN\cdot MP-MN(MN+MP)}{MP(MN+MP)}=\\ \\=\dfrac{MN\cdot MP-MN^2-MN\cdot MP}{MP(MN+MP)}=\dfrac{-MN^2}{MP(MN+MP)}$