Find P(100)

Algebra Level 5

$P(x)$ is a polynomial of degree 98 such that $P(K) = \dfrac{1}{K}$ for $K = 1, 2, 3, \ldots , 99$ . Then the value of $P(100)$ is:

\frac1{100!}

\frac1{50}

100 ! + 1

\frac1{100}

1 solution

Deepak Kumar
Apr 19, 2015

Let
r(x) = x (p(x) - 1/x) = x p(x) - 1 ......eqn(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)=>r(x) = c(x - 1)(x - 2)(x - 3) . . . (x - 99) .......eqn(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! =>r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99! ........eqn(3)
We can combine equations (1) and (3) and let x = 100 to find p(100)

Find P(100)

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