Pyramid Investigations 6 – Reversing The Pyramid

1+2+3+.............+(n-1)+n+(n-1)+...........+3+2+1 =289
2[1+2+3+.......+n]-n =289
n(n+1)-n = 289
n^2 = 289
n=17

$N+2K=289\\ K=\sum _{ i=1 }^{ N-1 }{ i } =\frac { (N-1)N }{ 2 } \\ N+{ N }^{ 2 }-N=289\\ N=\sqrt { 289 } =17 (+value)$

See the solution posted by me for this problem (link in the word "this") which depicts that the sum of such series of $n$ can be calculated by $n^2$ . So, from that we can rewrite the problem as -->

$n^2=289 \implies n=\boxed{17}$

We ignore $n=(-17)$ because the series cannot proceed with $n=(-17)$ .

We could observe a pattern:

$1=1$

$1+2+1=4$

$1+2+3+2+1=9$

$1+2+3+\ldots+(n-1)+n+(n-1)+\ldots+3+2+1=n^2$

$n^2=289 \Rightarrow n=\sqrt {289}=17$

The first part (1+2+3+4......(n-1)+n) is just the sum of first n natural numbers, n(n+1)/2. The last part((n-1)+....3+2+1) is the sum of the first (n-1) natural numbers, or n(n-1)/2. Thus 1+2+3...+(n-1)+n+(n-1)....+3+2+1=n(n+1)/2 + n(n-1)/2=n(n+1+n-1)/2=n^2. Thus n^{2}=289 or n=17.

Let $a = 1 + 2 + 3 + \ldots + (n - 1) + n + (n - 1) + \ldots + 3 + 2 + 1$

Given that

$\displaystyle \sum_{1}^x = \frac {x(x + 1)}{2}$

and

$a = 2(1 + 2 + 3 + \ldots + (n - 1)) + n$

meaning that

$a = 2\displaystyle \sum_{1}^{n - 1} + n$

Using the formula we get

$a = 2\frac {(n - 1)(n - 1 + 1)}{2} + n$

Simplifying this gives

$a = \frac {2n(n - 1)}{2} + n \Rightarrow n(n - 1) + n \Rightarrow n^2 - n + n \Rightarrow n^2$

Since $a = 289$ and $a = n^2$

$n^2 = 289 \Rightarrow n = \sqrt {289} \Rightarrow n = 17$

So the answer is $17$

$1 + 2 + 3 +... + (n-1) + n + (n-1) +...+ 3 + 2 + 1 = 289$

$2 (1 + 2 + 3 +... + (n-1))+n = 289$

$\{ 1+2+3+...+(n-1)\} \quad =\quad \frac { 289\quad -\quad n }{ 2 }$

$\frac { n-1 }{ 2 } (2+n-2)\quad =\quad \frac { 289\quad -\quad n }{ 2 }$

$(n-1)\times \quad n\quad =\quad 289\quad -\quad n$

$n^{ 2 }\quad -\quad n\quad =\quad 289$

$n^{ 2 }\quad =\quad 289$

n = 17

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16 = 136, 136 + 136 = 272, 289 - 272 = 17

it's a series prefer sum number. we can calculate this like n^2=289. for this 17^2=289 so answer is 17.