E.S. 2008 – 6

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

Osserva

  1. {n \choose k} =\frac{n!}{k!(n-k)!}
  2. {n \choose k} =\frac{n(n-1)\ \cdots\ (n-k+1)}{k!}
  3. n \ge k \ge 0
  4. n\ge1n\ge2n\ge3n \ge 3
  5. {n \choose 1}+x={n \choose 2}
  6. {n \choose 2}+x={n \choose 3}

Si tratta di risolvere l’equazione

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

{n \choose 3}-2{n \choose 2}+{n \choose 1}=0

n=7


In alternativa calcola i coefficienti binomiali

n {n \choose 1} {n \choose 2} {n \choose 3}
3 {3 \choose 1} = 3 {3 \choose 2} = 3 {3 \choose 3} = 1
4 {4 \choose 1} = 4 {4 \choose 2} = 6 {4 \choose 3} = 4
5 {5 \choose 1} = 5 {5 \choose 2} = 10 {5 \choose 3} = 10
6 {6 \choose 1} = 6 {6 \choose 2} = 15 {6 \choose 3} = 20
7 {7 \choose 1} = 7 {7 \choose 2} = 21 {7 \choose 3} = 35

oppure genera le combinazioni e contale…

n {n \choose 1} {n \choose 2} {n \choose 3}
3 a b c
3
ab ac bc
3
abc
1
4 a b c d
4
ab ac ad bc bd cd
6
abc abd acd bcd
4
5 a b c d e
5
ab ac ad ae bc bd be cd ce de
10
abc abd abe acd ace ade bcd bce bde cde
10
6 a b c d e f
6
ab ac ad ae af bc bd be bf cd ce cf de df ef
15
abc abd abe abf acd ace acf ade adf aef bcd bce bcf bde bdf bef cde cdf cef def
20
7 a b c d e f g
7
ab ac ad ae af ag bc bd be bf bg cd ce cf cg de df dg ef eg fg
21
abc abd abe abf abg acd ace acf acg ade adf adg aef aeg afg bcd bce bcf bcg bde bdf bdg bef beg bfg cde cdf cdg cef ceg cfg def deg dfg efg
35

e individua la progressione aritmetica

  • 7
  • 21 (7+14)
  • 35 (21+14)
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