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

In summary, the formula for the sum of the first n terms of an arithmetic sequence is Sn = (n/2)(a1 + an). Induction can be used to solve the problem of determining the value of n in a given equation by showing that the equation holds true for n = 1 and then using this assumption to prove it for all values of k+1. The value of n is 13 in the equation 1 + 3 + 6 + .... + n(n - 1)/2 = 364. Proving the equation with induction is important because it shows that it holds true for all values of n, not just the one we are trying to determine. Induction can also be applied to other mathematical
  • #1
Monoxdifly
MHB
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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 + ... + Sn = 364. However, by assuming that Sn = \(\displaystyle \frac{1}{2}\)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?
 
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  • #2
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:

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

\(\displaystyle \sum_{k=1}^n\left(k^2\right)\)
 
  • #3
MarkFL said:
What are the formulas for:

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

MarkFL said:
What are the formulas for:

\(\displaystyle \sum_{k=1}^n\left(k^2\right)\)
I don't know...
 
  • #4
Monoxdifly said:
\(\displaystyle \frac{1}{2}n(n-1)?\)

Not quite, it is:

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

Monoxdifly said:
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?
 
  • #5
Monoxdifly said:
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 + ... + Sn = 364. However, by assuming that Sn = \(\displaystyle \frac{1}{2}\)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.
 
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  • #6
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:

\(\displaystyle n=13\)
 

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

What is the formula for the sum of the first n terms of an arithmetic sequence?

The formula for the sum of the first n terms of an arithmetic sequence is Sn = (n/2)(a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term.

How can we use induction to solve the problem of determining the value of n in the given equation?

We can use induction to solve this problem by first showing that the equation holds true for n = 1. Then, we assume that it holds true for any value of k and use this assumption to prove that it also holds true for k+1. This will show that the equation holds true for all values of n, including the one we are trying to determine.

What is the value of n if 1 + 3 + 6 + .... + n(n - 1)/2 = 364?

The value of n is 13. This can be determined by plugging in different values of n into the equation until we find the one that equals 364.

What is the importance of proving the equation with induction rather than just solving for n algebraically?

Proving the equation with induction allows us to show that the equation holds true for all values of n, not just the one we are trying to determine. This gives us a more comprehensive understanding of the problem and its solution.

Can the concept of induction be applied to other mathematical problems?

Yes, induction can be applied to other mathematical problems. It is a powerful tool for proving equations and solving problems in various branches of mathematics, including algebra, calculus, and number theory.

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