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

  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Induction Value
AI Thread Summary
The discussion focuses on determining the value of n in the equation 1 + 3 + 6 + ... + (1/2)n(n - 1) = 364. Participants clarify that the sum of the series can be expressed using the formula for the sum of squares, leading to the equation (2n^3 - 2n - 4368 = 0). After applying the rational roots theorem, they find that n = 13 is a solution. The quadratic factor resulting from synthetic division has a negative discriminant, confirming that n = 13 is the only real solution. The final conclusion is that the value of n is 13.
Monoxdifly
MHB
Messages
288
Reaction score
0
Determine the value of n if 1 + 3 + 6 + ... + $$\frac{1}{2}$$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 + ... + Sn = 364. However, by assuming that Sn = $$\frac{1}{2}$$n(n - 1) I got n + 1 = n - 1 which is simply unsolvable at all. After all, the term $$\frac{1}{2}$$n(n - 1) doesn't match for n = 1. Does this question even have any solution?
 
Mathematics news on Phys.org
I would write:

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

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

What are the formulas for:

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

$$\sum_{k=1}^n\left(k^2\right)$$
 
MarkFL said:
What are the formulas for:

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

MarkFL said:
What are the formulas for:

$$\sum_{k=1}^n\left(k^2\right)$$
I don't know...
 
Monoxdifly said:
$$\frac{1}{2}n(n-1)?$$

Not quite, it is:

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

Monoxdifly said:
I don't know...

Now for:

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

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

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

Now, consider, that we have:

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

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

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

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

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

So, this gives rise to the system:

$$d=0$$

$$a+b+c=1$$

$$8a+4b+2c=5$$

$$27a+9b+3c=14$$

I would use Gaussian elimination to solve this system:

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

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

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

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

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

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

This tells us:

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

Hence:

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

Now, can you proceed with these formulas?
 
Monoxdifly said:
Determine the value of n if 1 + 3 + 6 + ... + $$\frac{1}{2}$$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 + ... + Sn = 364. However, by assuming that Sn = $$\frac{1}{2}$$n(n - 1) I got n + 1 = n - 1 which is simply unsolvable at all. After all, the term $$\frac{1}{2}$$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.
 
Last edited:
To follow up, we have:

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

Using our summation formulas, we get:

$$\frac{2n^3+3n^2+n}{6}-\frac{n^2+n}{2}=728$$

Multiply through by 6:

$$2n^3+3n^2+n-3n^2-3n=4368$$

Arrange in standard form:

$$2n^3-2n-4368=0$$

Divide through by 2:

$$n^3-n-2184=0$$

Let:

$$f(n)=n^3-n-2184$$

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

$$f(13)=0$$

Then, using synthetic division, we obtain:

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

This tells us:

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

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

$$n=13$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top