Problem:
Assume that x1,x2,…,x7 are real numbers such that
x1+4x2+9x3+16x4+25x5+36x6+49x7=4x1+9x2+16x3+25x4+36x5+49x6+64x7=9x1+16x2+25x3+36x4+49x5+64x6+81x7=112123.
Find the value of
16x1+25x2+36x3+49x4+64x5+81x6+100x7.
Solution:
Observe that subtracting 3 times the second equation from the sum of the first equation and 3 times the third equation gives an equation with the desired expression on the left. Thus
16x1+25x2+36x3+49x4+64x5+81x6+100x7=1(1)−3(12)+3(123)=334
More formally, we can deduce the relation mentioned above by finding constants a,b and c such that
an2+b(n+1)2+c(n+2)2=(n+3)2(1)
holds for all n. Considering (1) as a polynomial identity in n, we expand and simplify both sides, then equate coefficients of like powers. of n. We obtain the equations
a+b+c2b+4cb+4c=1=6=9
The solution of this system is a=1,b=−3 and c=3.
Note. For the sake of completeness, one should check that the system of three equations given in the problem does have a solution. One such solution is x1=797/4, x2=−916/4,x3=319/4 and x4=x5=x6=x7=0.
It is interesting to note that the number of variables and their values are of little significance. The reader may wish to investigate generalizations of these results to problems in which the coefficients are cubes, fourth powers, etc.
The problems on this page are the property of the MAA's American Mathematics Competitions