Question

Ma poate ajuta cineva?

Answer 1

Salut,

Punctul a):

$a_1+a_2+\ldots+a_{2014}=f(1)=(1+1)^{2014}=2^{2014}.$

Punctul b):

$\big{a}_0=C_{2014}^0=\dfrac{2014!}{(2014-0)!\cdot 0!}=1;\\\\\\\big{a}_{2014}=C_{2014}^{2014}=\dfrac{2014!}{(2014-2014)!\cdot 2014!}=1;\\\\\\\big{a}_{2013}=C_{2014}^{2013}=\dfrac{2014!}{(2014-2013)!\cdot 2013!}=\dfrac{2014\cdot 2013!}{2013!}=2014.$

Punctul c:

$f(-\sqrt2)\cdot f(\sqrt2)=(1-\sqrt2)^{2014}\cdot(1+\sqrt2)^{2014}=\left[(1-\sqrt2)\cdot(1+\sqrt2)\right]^{2014}=\\\\=\left\{[1^2-(\sqrt2)^2]\right\}^{2014}=(1-2)^{2014}=(-1)^{2014}=1.$

Green eyes.