Maths Olympiad Prep

Library / /399 of 520

Algebra Difficulty 6.0 AIME, harder Prove it

Someone used a table of cubes of natural numbers for squaring. They calculated the square of 2.5 according to the following scheme. Let's determine the procedure's schema, as well as whether the procedure is generally valid.

3.53=42.8753.5^{3}=42.875

1.53=3.3751.5^{3}=3.375

their difference: d=39.5\quad d=39.5

d−2=37.5d-2=37.5

1/61/6 of it: 6.25=2.52\quad 6.25=2.5^{2}.

Solution

The task does not specify how the individual formed the numbers 3.5 and 1.5 raised to the power of two. If these numbers were derived by adding 1 to 2.5 and subtracting 1 from 2.5, respectively, then the procedure is valid for any number x x , as shown by the following identity:

(x+1)3−(x−1)3−26=x2 \frac{(x+1)^{3}-(x-1)^{3}-2}{6}=x^{2}

Nagy Zoltán (Budapest, XI., Fehérvári úti 12-year school, 6th grade)

Note. A somewhat more general identity holds if k≠0 k \neq 0 :

(x+k)3−(x−k)3−2k36k=x2 \frac{(x+k)^{3}-(x-k)^{3}-2 k^{3}}{6 k}=x^{2}

Antos Péter (Budapest, Apáczai Csere J. secondary grammar school, 2nd grade)

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.