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Let $a,b$ be given positive integers such that $a<b$. Let $M(a,b)$ be defined as $$ \frac{\sum_{k=a}^b \sqrt{k^2+3k+3}}{b-a+1}. $$

Evaluate $[M(a,b)]$ where $[.]$ represents greatest integer function.

I have tried to factorise the term inside the square root however I believe it was pretty useless. I have a feeling that it might(?) telescope but I have no clue what to do here. Any help will be appreciated

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You should use the estimates $(x+1)^2<x^2+3x+3<(x+2)^2$. Then you get that $$\frac{\sum_{k=a}^b(k+1)}{b-a+1}<M(a,b)<\frac{\sum_{k=a}^b(k+2)}{b-a+1}$$ Using the fact that $\sum_{k=0}^nk=n(n+1)/2$ and the appropriate translations (write $k=k-a+a$ so you re-index the sums starting from $0$) we get the estimates $$1+\frac{b-a}{2}+a<M(a,b)< 2+\frac{b-a}{2}+a$$ This concludes:

If $b-a$ is even, then $[M(a,b)]=1+(b-a)/2+a$.

If $b-a$ is odd, then $[M(a,b)]=1/2+(b-a)/2+a$.

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    $\begingroup$ Thank you ! You just saved me from insanity. Appreciate it $\endgroup$
    – Samyak Jha
    Jun 11 '19 at 13:24
  • $\begingroup$ @SamyakJha No problem! $\endgroup$ Jun 11 '19 at 13:24

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