I tried to approximate $\displaystyle\bigg\lfloor \sum^{10}_{n=2}n^{\frac{1}{12-n}}\bigg\rfloor$ using inequality.
My try was,
$\sqrt[n-k+1]{k-1}<\sqrt[n-k]{k}<\sqrt[n-k-1]{k+1}$
So,
$7\sqrt[12]{2}<\displaystyle\sum^{10}_{n=2}n^{\frac{1}{12-n}}<7\sqrt{12}$
$\displaystyle 8\leq\bigg\lfloor \sum^{10}_{n=2}n^{\frac{1}{12-n}}\bigg\rfloor\leq24$
So, is there a way to approximate it more accurately?