## watchmath 5 years ago 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 $$16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$$

1. anonymous

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

2. watchmath

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

3. anonymous

yes less work too never thought of that nice :)