Esame di Stato 2010 – 8

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Se n>3 e \displaystyle {n \choose n-1}, \displaystyle {n \choose n-2}, \displaystyle {n \choose n-3}, sono in progressione aritmetica, qual è il valore di n?

Soluzione 1

Progressione aritmetica… quindi

\displaystyle {n \choose n-1}+k = {n \choose n-2}

\displaystyle {n \choose n-2}+k = {n \choose n-3}

Si tratta di risolvere l’equazione seguente, per n > 3

\displaystyle {n \choose n-1}-{n \choose n-2} = {n \choose n-2}-{n \choose n-3}

…

n=7

Soluzione 2

Fai i calcoli per n=4, 5, 6, … finché non riconosci la progressione 7, 7+14=21, 7+14+14=35

n\displaystyle {n \choose {n-1}}\displaystyle {n \choose {n-2}}\displaystyle {n \choose {n-3}}
4\displaystyle {4 \choose 3} = \displaystyle \frac{4!}{3!(4-3)!} = 4\displaystyle {4 \choose 2} = \displaystyle \frac{4!}{2!(4-2)!} = 6\displaystyle {4 \choose 1} = \displaystyle \frac{4!}{1!(4-1)!} = 4
5\displaystyle {5 \choose 4} = \displaystyle \frac{5!}{4!(5-4)!} = 5\displaystyle {5 \choose 3} = \displaystyle \frac{5!}{3!(5-3)!} = 10\displaystyle {5 \choose 2} = \displaystyle \frac{5!}{2!(5-2)!} = 10
6\displaystyle {6 \choose 5} = \displaystyle \frac{6!}{5!(6-5)!} = 6\displaystyle {6 \choose 4} = \displaystyle \frac{6!}{4!(6-4)!} = 15\displaystyle {6 \choose 3} = \displaystyle \frac{6!}{3!(6-3)!} = 20
7\displaystyle {7 \choose 6} = \displaystyle \frac{7!}{6!(7-6)!} = 7\displaystyle {7 \choose 5} = \displaystyle \frac{7!}{5!(7-5)!} = 21\displaystyle {7 \choose 4} = \displaystyle \frac{7!}{4!(7-4)!} = 35

Soluzione 3

Genera tutte le sequenze…, contale e fermati quando riconosci la progressione

n\displaystyle {n \choose {n-1}}\displaystyle {n \choose {n-2}}\displaystyle {n \choose {n-3}}
4ABC ABD ACD BCD4AB AC AD BC BD
CD		6A B C D4
5ABCD ABCE ABDE ACDE BCDE5ABC ABD ABE ACD ACE
ADE BCD BCE BDE CDE		10AB AC AD AE BC
BD BE CD CD DE		10
6ABCDE ABCDF ABCEF ABDEF ACDEF
BCDEF		6ABCD ABCE ABCF ABDE ABDF
ABEF ACDE ACDF ACEF ADEF
BCDE BCDF BCEF BDEF CDEF		15ABC ABD ABE ABF ACD
ACE ACF ADE ADF AEF
BCD BCE BCF BDE BDF
BEF CDE CDF CEF DEF		20
7ABCDEF ABCDEG ABCDFG ABCEFG ABDEFG
ACDEFG BCDEFG		7ABCDE ... CDEFG21ABCD ... DEFG35