Question

Repost:
Let \(x_1,\ldots,x_7\) be real numbers such that
\[\begin{align*}
x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7&=1\\
4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7&=12\\
9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7&=123
\end{align*}\]

Compute
\(16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7\)

Answer 1

\[\sum_{n=1}^{7}n ^{2}X _{n}=1\]
\[\sum_{n=1}^{7}(n+1)^{2}X _{n} = \sum_{n=1}^{7}n^{2}X _{n}+2nX _{n}+X _{n}=12\]
\[\sum_{n=1}^{7}(n+2)^{2}X _{n} = \sum_{n=1}^{7}n^{2}X _{n}+4nX _{n}+4X _{n}=123\]

\[u = \sum_{n=1}^{7}nX _{n}\]
\[v = \sum_{n=1}^{7}X _{n}\]
System of 2 equations
\[1+2u + v = 12\]
\[1+4u+4v = 123\]
solving yields
\[u = -\frac{39}{2}\]
\[v=50\]
\[\sum_{n=1}^{7}(n+3)^{2}X_n=1+6u+9v =334\]

Answer 2

Here is another way
\(n^2-3(n+1)^2+3(n+2)^2=(n+3)^2\) for all \(n\)
It follows that
\(16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=1-3(12)+3(123)=334\)

Answer 3

yes less work too
never thought of that
nice :)