Question

The sum of the first m squares of natural numbers is given by m(m+1)((2m+1) all over 6. for example 1^2+2^2+3^2+4^2=30 = 4(4+1)(2 x 4 +1) all over 6. Find all values of m such that lies between 300 and 100.

Answer 1

So basically, you have to find the values of m for which the sum of the all squares from 1 to m lie between 300 and 100. So let's set up an inequality perhaps?

So I've set up an inequality for the sum of squares to lie between 100 and 300. Now we have to somehow solve for m. \[\bf 100<\frac{ m(m+1)(2m+1) }{ 6 }<300\]Multiply all sides by 6 to eliminate the fraction.\[\bf 600<m(m+1)(2m+1)<1800\]So to solve this "compound inequality", we need to consider two cases. One for which m(m+1)(2m+1) is greater than 600 and one for which it's smaller than 1800. We will get the intervals of m which satisfy both conditions and take their intersection to find the values of that satisfy both constraints.

\(\bf Case \ 1:\) I will solve the case for the part that deals with " > 600 ". We will expand the factored expressed, bring the 600 over the other side, factor again and solve for the values of m.
\[\bf m(m+1)(2m+1)>600 \rightarrow 2m^3+3m^2+m-600>0\]\[\bf \implies m >6.207\]The expression cannot be factored normally so I obtained the root graphically.

\(\bf Case 2:\) Now we solve for the expression when it's smaller than 1800.\[\bf m(m+1)(2m+1)<1800 \rightarrow 2m^2+3m^2+m-1800<0\]\[\bf \implies m < 9.164\]So we have solved both cases, now we must find the intersection of both intervals of m from both cases to get the interval of m that satisfies both given constraints. The intersection of the two intervals we found is: \(\bf m \in (6.207,9164)\). And that's the interval for the values of m which satisfy the compound inequality.

Makes sense? If you got any questions, feel free to message/tag me. @isaiah95

~ genius12