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

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

Step-by-step explanation:

*face palms*

step-by-step explanation:

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