Tìm số nguyên dương n thỏa mãn: 1/2 Cn 0 - 1/3Cn 1 + 1/4Cn 2 - 1/5Cn 3

Tìm số nguyên dương n thỏa mãn:

$\frac{1}{2}C_n^0 - \frac{1}{3}C_n^1 + \frac{1}{4}C_n^2 - \frac{1}{5}C_n^3 + ... + \frac{{{{\left( { - 1} \right)}^n}}}{{n + 2}}C_n^n = \frac{1}{{156}}$.

Xét công thức tổng quát:

$\frac{{{{\left( { - 1} \right)}^k}}}{{k + 2}}C_n^k = \frac{{{{\left( { - 1} \right)}^k}}}{{k + 2}}\,.\,\frac{{n!}}{{k!\,.\,\left( {n - k} \right)!}} = \frac{{{{\left( { - 1} \right)}^k}\,.\,\left( {k + 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,\frac{{\left( {n + 2} \right)!}}{{\left( {k + 2} \right)!\,.\,\left( {n - k} \right)!}}$

$= \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,{\left( { - 1} \right)^k}\,.\,\left( {k + 1} \right)\,.\,C_{n + 2}^{k + 2}$

$= \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,\left[ {{{\left( { - 1} \right)}^k}\,.\,\left( { - 1} \right)\,.\,C_{n + 2}^{k + 2} + {{\left( { - 1} \right)}^k}\,.\,\left( {k + 2} \right)\,.\,C_{n + 2}^{k + 2}} \right]$

$= \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,\left[ {\left( { - 1} \right)\,.\,{{\left( { - 1} \right)}^{k + 2}}\,.\,C_{n + 2}^{k + 2} + {{\left( { - 1} \right)}^k}\,.\,\left( {n + 2} \right)\,.\,C_{n + 1}^{k + 1}} \right]$

$= - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,{\left( { - 1} \right)^{k + 2}}\,.\,C_{n + 2}^{k + 2} - \frac{1}{{n + 1}}\,.\,{\left( { - 1} \right)^{k + 1}}\,.\,C_{n + 1}^{k + 1}$

Khi đó: $\frac{1}{2}C_n^0 - \frac{1}{3}C_n^1 + \frac{1}{4}C_n^2 - \frac{1}{5}C_n^3 + ... + \frac{{{{\left( { - 1} \right)}^n}}}{{n + 2}}C_n^n$

$= - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\left[ {C_{n + 2}^2 - C_{n + 2}^3 + C_{n + 2}^4 + ... + {{\left( { - 1} \right)}^{n + 2}}C_{n + 2}^{n + 2}} \right]$

$- \frac{1}{{n + 1}}\left[ { - C_{n + 1}^1 + C_{n + 1}^2 - C_{n + 1}^3 + ... + {{\left( { - 1} \right)}^{n + 1}}C_{n + 1}^{n + 1}} \right]$

$= - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\left[ {{{\left( { - 1 + 1} \right)}^{n + 2}} - C_{n + 2}^0 + C_{n + 2}^1} \right] - \frac{1}{{n + 1}}\left[ {{{\left( { - 1 + 1} \right)}^{n + 1}} - C_{n + 1}^0} \right]$

\( =  - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,\left( { - 1 + n + 2} \right) - \frac{1}{{n + 1}}\,.\,\left( { - 1} \right)\)

$= - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}\,.\,\left( {n + 1} \right) - \frac{1}{{n + 1}}\,.\,\left( { - 1} \right)$

$= - \frac{1}{{n + 2}} + \frac{1}{{n + 1}} = \frac{{\left( {n + 2} \right) - \left( {n + 1} \right)}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

$= \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

Do đó, theo bải ra ta có: $\frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{156}}$

Û (n + 1)(n + 2) = 156

Û n² + 3n + 2 = 156

Û n² + 3n − 154 = 0

$\Leftrightarrow \left[ \begin{array}{l}n = 11\;\;\;\left( n \right)\\n = - 14\;\left( l \right)\end{array} \right.$

Vậy n = 11 là số nguyên dương cần tìm.