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\Sigma^{b}_{k=a}\sqrt{k^2+3k+3} /(b-a+1), floor function sum

Source: Austrian Federal Competition For Advanced Students 2009, Part 2, p2

August 31, 2019

algebrafloor functionfunction

Problem Statement

(i) For positive integers

a<b

, let

M(a,b)=\frac{\Sigma^{b}_{k=a}\sqrt{k^2+3k+3}}{b-a+1}

. Calculate

[M(a,b)]

(ii) Calculate

N(a,b)=\frac{\Sigma^{b}_{k=a}[\sqrt{k^2+3k+3}]}{b-a+1}