# Thread: cool matrix problem... Try this .. will be interesting

1. ## cool matrix problem... Try this .. will be interesting

Matrix M1=[1];
Matrix M2= 2 3
( 2*2 ) 4 5

Matrix (3*3)M3 = 6 7 8
9 10 11
12 13 14

Matrix(4*4) M4= 15 16 17 18
19 20 21 22
23 24 25 26
27 28 29 30

..... etc for Matrix Mn

Find the sum of elemnts of the main diagonal. and give the solution in terms of A+Bn+Cn^2...

2. Originally Posted by kamaksh_ice
Matrix M1=[1];
Matrix M2= 2 3
( 2*2 ) 4 5

Matrix (3*3) M3 = 6 7 8
9 10 11
12 13 14

Matrix(4*4) M4 = 15 16 17 18
19 20 21 22
23 24 25 26
27 28 29 30

..... etc for Matrix Mn.

Find the sum of elemnts of the main diagonal. and give the solution in terms of A+Bn+Cn^2...
So you are asking for the general term of the series $\displaystyle \{1, 8, 38, ... \}$ ?

3. Hello, kamaksh_ice!

$\displaystyle M_1\:=\:\begin{pmatrix}1\end{pmatrix}\quad M_2\:=\:\begin{pmatrix}2 & 3 \\ 4 & 5\end{pmatrix} \quad M_3\:=\:\begin{pmatrix}6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14\end{pmatrix} \quad M_4 \:=\:\begin{pmatrix}15 & 16 & 17 & 18 \\ 19 & 20 & 21 & 22 \\ 23 & 24 & 25 & 26 \\ 27 & 28 & 29 & 30\end{pmatrix}$
... and so on to matrix $\displaystyle M_n$.

Find the sum of elements of the main diagonal.

We have: .$\displaystyle \begin{array}{cccc} 1 & = & 1 & [1] \\ 2 + 5 & = & 7 & [2]\\ 6 + 10 + 14 & = & 30 & [3]\\ 15 + 20 + 25 + 30 & = & 90 & [4]\\ 31 + 37 + 43 + 49 + 55 & = & 215 & [5]\end{array}$

Each is the sum of an arithmetic series.

. . $\displaystyle \begin{array}{cccc}\text{In [3]:} & a = 6 & d = 4 & n = 3 \\ \text{In [4]:} & a = 15 & d = 5 & n = 4 \\ \text{In [5]:} & a = 31 & d = 6 & n = 5\end{array}$

In the $\displaystyle n^{th}$ matrix, the common difference is $\displaystyle n+1$, and there are $\displaystyle n$ terms.

We must determine the first terms of these series: $\displaystyle 1,\,2,\,6,\,15,\,31,\,56\cdots$

After some algebra, we find that: .$\displaystyle a \:=\:\frac{1}{6}(n+1)(2n^2-5n+6)$

The sum of an arithmetic series is: .$\displaystyle S \;=\;\frac{n}{2}[2a + (n-1)d]$

So we have: .$\displaystyle S \;=\;\frac{n}{2}\left[2\cdot\frac{1}{6}(n+1)(2n^2-5n+6) + (n-1)(n+1)\right]$

. . which simplifies to: .$\displaystyle \boxed{S \;=\;\frac{n}{6}(n+1)(2n^2-2n+3)}$