Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that
\begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*}
Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$.
Note that each given equation is of the form \[f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7\] for some $k\in\{1,2,3\}.$
When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ \[f(k)=ak^2+bk+c,\] where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$
We are given that
\begin{alignat*}{10} f(1)&=\phantom{42}a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123, \end{alignat*}
and we wish to find $f(4).$
We eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third:
\begin{align*} 3a+b&=11, \\ 5a+b&=111. \end{align*}
By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$
Finally, the answer is \[f(4)=16a+4b+c=\boxed{334}.\]
~Azjps ~MRENTHUSIASM