[ASK] Induction Determine the value of n if 1 + 3 + 6 + .... + n(n - 1)/2 = 364

[Monoxdifly]

Determine the value of n if 1 + 3 + 6 + ... + \(\displaystyle \frac{1}{2}\)n(n - 1) = 364

What I did:
I know that 1, 3, and 6 are the result of arithmetic series with the starting value 1 and the difference 2, thus that sum can be written as S₁ + S₂ + S₃ + ... + S_n = 364. However, by assuming that S_n = \(\displaystyle \frac{1}{2}\)n(n - 1) I got n + 1 = n - 1 which is simply unsolvable at all. After all, the term \(\displaystyle \frac{1}{2}\)n(n - 1) doesn't match for n = 1. Does this question even have any solution?

 

[MarkFL]

I would write:

\(\displaystyle \frac{1}{2}\sum_{k=1}^n\left(k(k-1)\right)=364\)

\(\displaystyle \sum_{k=1}^n\left(k^2-k\right)=728\)

What are the formulas for:

\(\displaystyle \sum_{k=1}^n\left(k\right)\)

\(\displaystyle \sum_{k=1}^n\left(k^2\right)\)

 

[Monoxdifly]

What are the formulas for:

\(\displaystyle \sum_{k=1}^n\left(k\right)\)

\(\displaystyle \frac{1}{2}n(n-1)?\)

What are the formulas for:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)\)

I don't know...

 

[MarkFL]

\(\displaystyle \frac{1}{2}n(n-1)?\)

Not quite, it is:

\(\displaystyle \sum_{k=1}^n\left(k\right)=\frac{n(n+1)}{2}\)

I don't know...

Now for:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)\)

The sum of squared terms is going to be a cubic function, so let's write:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)=an^3+bn^2+cn+d\)

Now, consider, that we have:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)=\sum_{k=0}^n\left(k^2\right)\)

\(\displaystyle \sum_{k=0}^1\left(k^2\right)=0^2=0\)

\(\displaystyle \sum_{k=1}^1\left(k^2\right)=0+1^2=1\)

\(\displaystyle \sum_{k=1}^2\left(k^2\right)=1+2^2=5\)

\(\displaystyle \sum_{k=1}^3\left(k^2\right)=5+3^2=14\)

So, this gives rise to the system:

\(\displaystyle d=0\)

\(\displaystyle a+b+c=1\)

\(\displaystyle 8a+4b+2c=5\)

\(\displaystyle 27a+9b+3c=14\)

I would use Gaussian elimination to solve this system:

\(\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 1 \\ 8 & 4 & 2 & 5 \\ 27 & 9 & 3 & 14 \end{array}\right]\)

\(\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 1 \\ 0 & -4 & -6 & -3 \\ 0 & -18 & -24 & -13 \end{array}\right]\)

\(\displaystyle \left[\begin{array}{ccc|c}1 & 1 & 1 & 1 \\ 0 & 1 & \frac{3}{2} & \frac{3}{4} \\ 0 & -18 & -24 & -13 \end{array}\right]\)

\(\displaystyle \left[\begin{array}{ccc|c}1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ 0 & 1 & \frac{3}{2} & \frac{3}{4} \\ 0 & 0 & 3 & \frac{1}{2} \end{array}\right]\)

\(\displaystyle \left[\begin{array}{ccc|c}1 & 0 & -\frac{1}{2} & \frac{1}{4} \\ 0 & 1 & \frac{3}{2} & \frac{3}{4} \\ 0 & 0 & 1 & \frac{1}{6} \end{array}\right]\)

\(\displaystyle \left[\begin{array}{ccc|c}1 & 0 & 0 & \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{6} \end{array}\right]\)

This tells us:

\(\displaystyle (a,b,c,d)=\left(\frac{1}{3},\frac{1}{2},\frac{1}{6},0\right)\)

Hence:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n=\frac{2n^3+3n^2+n}{6}=\frac{n\left(2n^2+3n+1\right)}{6}=\frac{n(n+1)(2n+1)}{6}\)

Now, can you proceed with these formulas?

 

[MountEvariste]

Determine the value of n if 1 + 3 + 6 + ... + \(\displaystyle \frac{1}{2}\)n(n - 1) = 364

What I did:
I know that 1, 3, and 6 are the result of arithmetic series with the starting value 1 and the difference 2, thus that sum can be written as S₁ + S₂ + S₃ + ... + S_n = 364. However, by assuming that S_n = \(\displaystyle \frac{1}{2}\)n(n - 1) I got n + 1 = n - 1 which is simply unsolvable at all. After all, the term \(\displaystyle \frac{1}{2}\)n(n - 1) doesn't match for n = 1. Does this question even have any solution?

I think you maybe confusing two things here. If you say the sum can be written as $S_{1}+S_{2}+\cdots+S_{n} = 364$, then $S_n$ is the $n^{th}$ term of the sequence which, as you say, is $\frac{1}{2} n(n-1)$. This is different from the sum. The question is asking you to find $n$ if the sum $S_{1}+S_{2}+\cdots+S_{n}$ evaluates to $364$. Maybe the choice of naming the terms $S_1, S_2, \ldots, S_n$ is a cause for confusion as you might be used to denoting the value of the sum as $S_n$, in which case perhaps naming them $T_1, T_2, \ldots, T_n$ might be better.

 

[MarkFL]

To follow up, we have:

\(\displaystyle \sum_{k=1}^n\left(k^2-k\right)=728\)

Using our summation formulas, we get:

\(\displaystyle \frac{2n^3+3n^2+n}{6}-\frac{n^2+n}{2}=728\)

Multiply through by 6:

\(\displaystyle 2n^3+3n^2+n-3n^2-3n=4368\)

Arrange in standard form:

\(\displaystyle 2n^3-2n-4368=0\)

Divide through by 2:

\(\displaystyle n^3-n-2184=0\)

Let:

\(\displaystyle f(n)=n^3-n-2184\)

Trying the factors of 2184 in according with the rational roots theorem, we find:

\(\displaystyle f(13)=0\)

Then, using synthetic division, we obtain:

\(\displaystyle \begin{array}{c|rr}& 1 & 0 & -1 & -2184 \\ 13 & & 13 & 169 & 2184 \\ \hline & 1 & 13 & 168 & 0 \end{array}\)

This tells us:

\(\displaystyle f(n)=(n-13)\left(n^2+13n+168\right)\)

We see that the discriminant of the quadratic factor is negative, leaving only:

\(\displaystyle n=13\)