Trust me it's the pattern

Since $(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$ , the last equation is the first minus 3 times the second plus three times the third, $4-3*44+3*444=\boxed{1204}$

Assume $P(n)=n^2x_1+...+(n+6)^2x_7$ . P(n) is a quadratic polynomial in $n$ i.e. $P(n)=an^2+bn+c$ . We are given $P(1)=4,P(2)=44,P(3)=444$ hence, we can find $a, b, c$ . We have to evaluate $P(4)$ which can be done after finding $a, b, c$ .

For an alternative solution, hehe I just investigated the pattern, as what the question suggested.

I subtracted equation (1) from equation (2) and arrived at

$E_1 = 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 +15 x_7 = 40$

Similarly, I subtracted equation (2) from equation (3) and arrived at

$E_2 = 5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 +17x_7 = 400$

Notice that every corresponding coefficient is two more than its counterpart, or

$Q_1 = E_2 - E1$

$Q_1= 2(x_1 + x_2 + x_3 + x_4 + x_5 + x_6 +x_7) = 360$

To arrive at the next pattern,we need to add $E_3$ to equation (3) such that

$E_3 = E_2 + Q_1$

$E_3 = 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 +19x_7 = 760$

so, we just add $E_3$ to equation (3) and we arrive at the equation above, with the RHS equal to

$760 + 444 = \boxed {1204}$ .

I solved for $x_1+x_2+x_4+......x_7$ but that was no need. Just need to multiply and subtract some equations.

The 3 given equations can be written as:

$\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { \ x }_{ i }$ =4 .....Equation 1

$\sum _{ i=1 }^{ 7 }{ { (i+1) }^{ 2 } } \times { \ x }_{ i }$ =44 .....Equation 2

$\sum _{ i=1 }^{ 7 }{ { (i+2) }^{ 2 } } \times { \ x }_{ i }$ =444 .....Equation 3

Now,

$\sum _{ i=1 }^{ 7 }{ { (i+1) }^{ 2 } } \times { \ x }_{ i }$ = $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }+1\times { x }_{ i }+2i\times { x }_{ i }$ = $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }\quad +\sum _{ i=1 }^{ 7 }{ 2i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ 1\times } { x }_{ i }$ =44 ....Equation 4

From Equation 1 , Equation 4 is equal to

$4+2\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ { x }_{ i } }$ =44

$\Rightarrow$ $2\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ { x }_{ i } }$ =40 ...Equation 5

Again,

$\sum _{ i=1 }^{ 7 }{ { (i+2) }^{ 2 } } \times { \ x }_{ i }$ = $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }+4\times { x }_{ i }+4i\times { x }_{ i }$ = $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }\quad +\sum _{ i=1 }^{ 7 }{ 4i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ 4\times } { x }_{ i }$ =444 ......Equation 6

From Equation 1 ,Equation 6 is equal to

$4+4\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }+4\sum _{ i=1 }^{ 7 }{ { x }_{ i } }$ =444

$\Rightarrow$ $\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ { x }_{ i } }$ =110..Equation7

Equation 5 - Equation 7

$\Rightarrow$ $\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }$ =-70 .....Equation 8

$\Rightarrow$ $\sum _{ i=1 }^{ 7 }{ { x }_{ i } }$ =180 ...Equation 9

The expression's value which we have to find can be written as

$\sum _{ i=1 }^{ 7 }{ { (i+3) }^{ 2 } } \times { \ x }_{ i }$

which is equal to $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }+9\times { x }_{ i }+6i\times { x }_{ i }$

$\Rightarrow$ $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }\ +\sum _{ i=1 }^{ 7 }{ 6i\times } { x }_{ i }+\sum _{ i=1 }^{ 7 }{ 9\times } { x }_{ i }$ = $\sum _{ i=1 }^{ 7 }{ { i }^{ 2 } } \times { x }_{ i }\ +6\sum _{ i=1 }^{ 7 }{ i\times } { x }_{ i }+9\sum _{ i=1 }^{ 7 }{ \times } { x }_{ i }$

Now the value can easily be found from equation 8,9 and 1

$\sum _{ i=1 }^{ 7 }{ { (i+3) }^{ 2 } } \times { \ x }_{ i }$ =4+ $6\times -70$ + $9\times 180$

$\therefore$ $16{ x }_{ 1 }+25{ x }_{ 2 }+36{ x }_{ 3 }+49x_{ 4 }+64{ x }_{ 5 }+81{ x }_{ 6 }+100{ x }_{ 7 }$ =4+1620-420= $\boxed{1204}$

Treating x4 = x5 = x6 = x7 = 0, solved by 4 Cramer's determinants:

x1 = 717

x2 = -824

x3 = 287

16 x1 + 25 x2 + 36 x3 = 1204

The value of expression wanted must be specially related to all above even though not necessarily sticks to (4, 44 , 444) like someone here had proposed with (16, 25, 36) for (x1, x2, x3) = (1, -3, 3) and there ought to be at least a valid case such as for x4 to x7 equal to zero!

I have done this in different way. It is clear that it is a linear system of equations. I will make a new system of equation to find out the multipliers by the coefficient of any three variables. Lets assume that the first equation is multiplied by X second by Y and third by Z. Then it can be written: $X + 4Y + 9Z=16, \\ 4X+9Y+16Z=25 \\ 9X+16Y + 25Z=36$ solve this system of equation and you will get: $X=1, \\ Y=-3, \\ Z=3$ hence now the multipliers are clear, apply the same rule for the row operation of the given system of equation and you will get $4 \times 1 + 44 \times -3 + 444 \times 3 = 1204$