Question

Coeficientii binomiali ai termenilor 5,6 si 7 din dezvoltarea $(1+x)^{n}$ sunt in progresie aritmetica.Aflati n.

Answer 1

Salut,

În general:

$T_{k+1}=C_n^k\cdot a^{n-k}\cdot b^k;\\\\T_{5}=C_n^4\cdot 1^{n-4}\cdot x^4;\\\\T_{6}=C_n^5\cdot 1^{n-5}\cdot x^5;\\\\T_{7}=C_n^6\cdot 1^{n-6}\cdot x^6;\\\\Coeficien\c{t}ii\;binomiali\;sunt:\;C_n^4,\;C_n^5,\;C_n^6.\\\\2\cdot C_n^5=C_n^4+C_n^6,\;sau\;2\cdot\dfrac{n!}{5!\cdot(n-5)!}=\dfrac{n!}{4!\cdot(n-4)!}+\dfrac{n!}{6!\cdot(n-6)!},\\\\\dfrac{2\cdot n!}{120\cdot(n-5)(n-6)!}=\dfrac{n!}{24\cdot(n-4)(n-5)(n-6)!}+\dfrac{n!}{6!\cdot(n-6)!}\Bigg|\cdot\dfrac{(n-6)!}{n!}\Rightarrow\\\\\Rightarrow\dfrac{1}{60(n-5)}=\dfrac{1}{24(n-4)(n-5)}+\dfrac1{720}$

Te las pe tine să continui, vei obţine că n=7 şi n=14. Spor la treabă !

Green eyes.