Esame di Stato 2014 – 3

Nello sviluppo di $\displaystyle \left(2a^2-3b^3 \right)^n$ compare il termine $\displaystyle -1080a^4b^9$ .
Qual è il valore di n?

Osserva

$\displaystyle (a+b)^n$ = $\displaystyle \sum_{k=0}^{n}{n \choose k}a^{(n-k)} b^{k}$
$\displaystyle \left(2a^2-3b^3\right)^n$ = $\displaystyle \sum_{k=0}^{n}{n \choose k}\left(2a^2\right)^{n-k}\left(-3b^3 \right)^{k}$
$\displaystyle {n \choose k}\left(2a^2\right)^{n-k}\left(-3b^3 \right)^{k}$ = $\displaystyle -1080a^4b^9$

Quindi

$\displaystyle \left(b^3\right)^k$ = $\displaystyle b^9$ , $\displaystyle 3k = 9$ , $\displaystyle k = 3$
$\displaystyle \left(a^2\right)^{n-k}$ = $\displaystyle a^4$ , $\displaystyle 2(n-k) = 4$ , $\displaystyle 2(n-3) = 4$ , $\displaystyle n = 5$

Infatti

$\displaystyle {5 \choose 3}\left(2a^2\right)^{5-3}\left(-3b^3 \right)^{3}$ = $\displaystyle 10\cdot 4a^4\cdot \left(-27b^9\right)$ = $\displaystyle -1080a^4b^9$