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If p is the smallest of four integers, what is their sum interms of P​

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Answer:

If p is the smallest of n consecutive integers of the same sign than we have p , p+1 , p+2 , … , p+(n−1) ,

So the sum is

∑k=0n−1(p+k)=∑k=0n−1p+∑k=0n−1k=np+n2−n2

Here n=4

So we have 4p+6

And checking

p+(p+1)+(p+2)+(p+3)=4p+6

Note if p=−v

Than you have the same thing as if p=v−n+1 just negative for example 3 consecutive integers the smallest is −5 so the sum is −5+(−4)+(−3)=3×−5+32−32=−15+3=−12

On the other hand:

−(3+4+5)=−(3×3+32−32)=−(9+3)=−12

If p=−v the sum of next v+1 integers is −(∑k=0vk)=−(v2+v2)

Than needs an other v integers to bring it up to 0 again. From there it is

∑k=0hk=h2+h2

Where h=n−(2v+1) .

So recap if p is the smallest of n consecutive integers their sum is

p+(p+1)+(p+2)+…+(p+(n−1))=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪np+n2+n2−(n|p|+n2+n2)((n−|p|)2+(n−|p|)2)−(|p|p+|p|2+|p|2)((n−|p|)2+(n−|p|)2)n≥0p<0∧n<|p|+1p<0∧|p|<n<2|p|+1p<0∧n>2|p|+1

Explanation:

User Alex Lobakov
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