2.71
Ta có
$C_n^0 + C_n^1 + C_n^2 = 79$
$<-> 1 + \dfrac{n!}{(n-1)!} + \dfrac{n!}{2! (n-2)!} = 79$
$<-> n + \dfrac{n(n-1)}{2} = 78$
$<-> 2n + n^2 - n = 156$
$<-> n^2 + n - 156 = 0$
Vậy $n = 12$ hoặc $n = -13$.
Do đó $n = 12$.
2.72
Ta có
$C_{n+4}^{n+1}- C_{n+3}^n = 7(n+3)$
$<-> \dfrac{(n+4)!}{(n+1)! 3!} - \dfrac{(n+3)!}{n! 3!} = 7(n+3)$
$<-> \dfrac{(n+2)(n+3)(n+4)}{6} - \dfrac{(n+1)(n+2)(n+3)}{6} = 7(n+3)$
$<-> (n+2)(n+3)(n+4) - (n+1)(n+2)(n+3) = 42(n+3)$
$<-> (n+3)[(n+2)(n+4) - (n+1)(n+2) - 42] = 0$
Ta có n + 3 > 0 nên ptrinh trở thành
$(n+2)(n+4) - (n+1)(n+2) - 42 = 0$
$<-> n^2 + 6n + 8 - (n^2 + 3n + 2) - 42 = 0$
$<-> 3n -36 = 0$
$<-> n = 12$
Vậy $n = 12$.
