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Algebra Difficulty 5.2 AIME, harder Find the answer

For 1j20141 \leq j \leq 2014, define bj=j2014i=1,ij2014(i2014j2014)b_{j}=j^{2014} \prod_{i=1, i \neq j}^{2014}(i^{2014}-j^{2014}) where the product is over all i{1,,2014}i \in\{1, \ldots, 2014\} except i=ji=j. Evaluate 1b1+1b2++1b2014\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We perform Lagrange interpolation on the polynomial P(x)=1P(x)=1 through the points 12014,22014,,201420141^{2014}, 2^{2014}, \ldots, 2014^{2014}. We have 1=P(x)=j=12014i=1,ij2014(xi2014)i=1,ij2014(j2014i2014)1=P(x)=\sum_{j=1}^{2014} \frac{\prod_{i=1, i \neq j}^{2014}(x-i^{2014})}{\prod_{i=1, i \neq j}^{2014}(j^{2014}-i^{2014})}. Thus, 1=P(0)=j=12014((1)2013)2014!2014j2014(1)2013i=1,ij2014(i2014j2014)1=P(0)=\sum_{j=1}^{2014} \frac{((-1)^{2013}) \frac{2014!^{2014}}{j^{2014}}}{(-1)^{2013} \prod_{i=1, i \neq j}^{2014}(i^{2014}-j^{2014})} which equals 2014!2014j=120141j2014i=1,ij2014(i2014j2014)=2014!2014(1b1+1b2++1b2014)2014!^{2014} \sum_{j=1}^{2014} \frac{1}{j^{2014} \prod_{i=1, i \neq j}^{2014}(i^{2014}-j^{2014})}=2014!^{2014}\left(\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{2014}}\right) so the desired sum is 12014!2014\frac{1}{2014!^{2014}}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.