Part 11

Rewrite this polynomial: $(x+1)^{2016}+(-x-2)^{2016}\geq (\frac{x-x-2+1}{2})^{2016}=\frac{1}{2^{2016}}=2^{-2015}$ The equality hold when: $x+1=-x-2$ $\Leftrightarrow x=-1.5$

It is given that $(x+1)^{2016} + (x+2)^{2016} = 2^{-2015}$

Let us consider the minimum of the LHS using AM-GM inequality, we have $(x+1)^{2016} + (x+2)^{2016} \ge 2(x+1)^{1008}(x+2)^{1008}$ .

And equality occurs when $(x+1)^{2016} = (x+2)^{2016}$ . Since $x+1 \ne x + 2$ , we consider:

$\begin{aligned} (x+1)^2 & = (x+2)^2 \\ x^2 + 2x + 1& = x^2 +4x + 4 \\ 2x & = -3 \\ \implies x & = - \frac{3}{2} \end{aligned}$

And the minimum value is $2(x+1)^{1008}(x+2)^{1008} = 2 \left( - \frac{1}{2} \right)^{1008} \left( - \frac{1}{2} \right)^{1008} = 2^{-2015}$ , which is equal to the RHS. Therefore, $x = \boxed{-\frac{3}{2}}$ for $(x+1)^{2016} + (x+2)^{2016} = 2^{-2015}$ .

Assume $\alpha$ = $x+1.5$ .

Then put value of $x$ in terms of $\alpha$ .

Then by expanding through binomial theoram and using its properties we find

$^{2016} C _0 (\alpha)^{2016} + ^{2016} C_1 (\alpha)^{2015} (0.5)........0.5^{2016}= 0.5^{2016}$ .

Clearly $\alpha=0$ .

That is $x= -1.5$ .