An algebra problem by Priyabrata Chowdhury
Algebra
Level
2
Let p(x) be a polynomial with degree 98 such that p(n) = 1/n for n = 1, 2, 3, 4, . . . , 99. Determine p(100).
1/71
9/12
1/21
1/50
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1 solution
Let
r(x) = x (p(x) - 1/x) = x p(x) - 1. (1) Since p(x) is a polynomial with degree 98, r(x) is a polynomial with degree 99. Since r(x) = x (p(x) - 1/x), and we are given that (p(x) - 1/x) = 0 for x = 1, 2, 3, . . . , 99,
r(x) has roots 1, 2, . . . , 99.
Since r(x) has degree 99, these are the only roots of r(x), which must thus have the form
r(x) = c(x - 1)(x - 2)(x - 3) . . . (x - 99) (2) for some constant c. To find c, we first let x = 0 in equation (1), yielding r(0) = -1. Letting x = 0 in (2) yields r(0) = -c(99!); hence, c = 1/99!. Thus, we have
r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99! (3)
We can combine equations (1) and (3) and let x = 100 to find
100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) . . . (100 - 99)/99! 100p(100) - 1 = 99!/99! = 1 p(100) = 1/50.