Baltuilhe wrote:Good afternoon!

This kind of problem is a Arithmetic Progression, 3rd order.

You can solve using the following idea:

[tex]\begin{array}{r|r|r|r}

\hline d_0&d_1&d_2&d_3\\\hline 1^2=1\\1+3^2=1+9=10&10-1=9\\10+5^2=10+25=35&35-10=25&25-9=16\\35+7^2=35+49=84&84-35=49&49-25=24&24-16=8\\84+9^2=84+81=165&165-84=81&81-49=32&32-24=8\\\hline\end{array}[/tex]

Now, the formula:

[tex]d_0\cdot\binom{n-1}{0}+d_1\cdot\binom{n-1}{2}+d_2\cdot\binom{n-1}{3}+d_3\cdot\binom{n-1}{4}\\\\

1\cdot\binom{n-1}{0}+9\cdot\binom{n-1}{2}+16\cdot\binom{n-1}{3}+8\cdot\binom{n-1}{4}\\\\

1\cdot\frac{(n-1)!}{0!(n-1)!}+9\cdot\frac{(n-1)!}{1!(n-1-1)!}+16\cdot\frac{(n-1)!}{2!(n-1-2)!}+8\cdot\frac{(n-1)!}{3!(n-1-3)!}\\\\

1\cdot 1+9\cdot(n-1)+16\cdot\frac{(n-1)(n-2)}{2}+8\cdot\frac{(n-1)(n-2)(n-3)}{6}\\\\

\frac{4n^3-n}{3}[/tex]

Now, the term [tex]1^2+2^2+3^2+\cdots+79^2[/tex] has [tex]\frac{79+1}{2}=40[/tex] terms. So:

[tex]\frac{4(40)^3-40}{3}=\frac{40(4(40)^2-1)}{3}=\frac{40(6\,400-1)}{3}=85\,320[/tex]

I hope to have helped!

@Baltuilhe

Thank you for the reply. Your insight really helped me understand how to solve this exercise and additionaly learn more about arithmetic sequences, I didn't know you could find the number of terms by adding the extremes and dividing by two You have my gratitude.

I have a couple questions though, Could you please answer when or If you are available to do so? The questions are:

1)The way you solved the question sort of reminds me of content related to matrices and combinatorics, does it have any relation to it?

2)May I have the name of the formula That you used after you wrote ''Now, the formula:'' or something to refer to it? Im quite interested in learning more about it.

3) Can I apply this concept to calculate the sum of any n order arithmetic sequence?

Once again You have my gratitude.

I appreciate your help.