Pyramid Investigations 4 – Pyramid Number Identity

We use here the formula $1+2+3+....+n=n+(n-1)+......+3+2+1=\frac{n(n+1)}{2}$ .

The given expression is ---->

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

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

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

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

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

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

$=(n-1)n+n = n((n-1)+1)=n(n-1+1)=n\times n = \boxed{n^2}$

always mid term square

1+2+3+4+..........+(n-1)+n+(n-1)+................+4+3+2+1
=[1+2+3+............+n]+[(n-1)+...........+4+3+2+1]
=[n(n+1)/2]+[(n-1)(n-1+1)/2]
= [n(n+1)/2]+[n(n-1)/2]
= (n/2)(n+1+n-1)=(n/2)(2n) = n^2

Its simplyer i think

n(n+1)/2+(n-1)n/2=n^2

The solution was available in a previous problem. I just remembered. Thank to whoever posted it.

A simpler way to explain is to write the series as follows:: Consider n=9., then you can rearrange the series as : 1+8+ 2+7+ 3+6+ 4+5+ 5+4+ 6+3+ 7+2+ 8+1+ 9 =9*9=9^2=n^. Here Answer is 81

This is getting stupid...n^n.

just we take square the numbers in squenvize and get ans in this way

square of n

1+2+3+......+(n-1)+n+(n-1)+......+3+2+1 =2(1)+2(2)+2(3)+.......+2(n-1)+n=n^2

Expression is from 1 to n and then back to 1 ,so sum will be nXn = n square

K.K.GARG,India

hint:use n=3

n^2=sum of all total sphere