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

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dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.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 \[u = \frac{39}{2}\] \[v=50\] \[\sum_{n=1}^{7}(n+3)^{2}X_n=1+6u+9v =334\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Here is another way \(n^23(n+1)^2+3(n+2)^2=(n+3)^2\) for all \(n\) It follows that \(16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=13(12)+3(123)=334\)

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.1yes less work too never thought of that nice :)
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