Question

În dezvoltarea binomului $( \sqrt{2^x} + \sqrt{2^{1-x} } )^n$ , n ∈ N*, suma coeficienților ultimilor trei termeni este egala cu 22. Să se afle valorile lui x pentru care suma dintre termenii al treilea si al cincilea este egală cu 135.

Answer 1

$\Big(\sqrt{2^x}+\sqrt{2^{1-x}}\Big)^n,\quad n\in \mathbb{N}^* \\ \\ C_{n}^{n-2}+C_{n}^{n-1}+C_{n}^n = 22 \\ \\ \dfrac{n!}{(n-2)!\cdot 2!} + \dfrac{n!}{(n-1)!\cdot 1!} +1 = 22 \\ \\ \dfrac{(n-1)\cdot n}{2}+ n = 21 \\n(n-1)+2n = 42\\ n(n-1+2) = 42 \\ \\ n(n+1) = 42 \Rightarrow n = 6$

$\Big(\sqrt{2^x}+\sqrt{2^{1-x}}\Big)^6 = \left(2^{\frac{x}{2}} + 2^{\frac{1-x}{2}}\right)^6 \\ \\ T_{k+1} = C_{n}^k\cdot a^{n-k}\cdot b^k \\ \\ T_{2+1} + T_{4+1} = 135\\ \\ C_{6}^2\cdot \Big(2^{\frac{x}{2}}\Big)^{6-2}\cdot \Big(2^{\frac{1-x}{2}}\Big)^2 +C_{6}^4\cdot \Big(2^{\frac{x}{2}}\Big)^{6-4}\cdot \Big(2^{\frac{1-x}{2}}\Big)^4 = 135$

$C_{6}^2\cdot \Big(2^{\frac{x}{2}}\Big)^{6-2}\cdot \Big(2^{\frac{1-x}{2}}\Big)^2 +C_{6}^4\cdot \Big(2^{\frac{x}{2}}\Big)^{6-4}\cdot \Big(2^{\frac{1-x}{2}}\Big)^4 = 135\\ \\ 15\cdot 2^{2x} \cdot 2^{1-x} + 15\cdot 2^{x}\cdot 2^{2(1-x)} = 135 \\ \\ 15\cdot 2^{2x+1-x}+15\cdot 2^{x+2-2x} = 135 \\\\ 15\cdot 2^{x+1}+15\cdot 2^{-x+2} = 135\\ \\ 30\cdot 2^{x} + 60\cdot 2^{-x} = 135 \\ \\ 2^{x} = t,\quad t>0\\ \\ 30t+60\cdot \dfrac{1}{t} = 135 \Big|\cdot t \\ \\ 30t^2+60 - 135t = 0 \\ \\30t^2-135t+60 = 0\Big|:5$
$6t^2-27t+12 = 0\Big|:3 \\ \\ 2t^2-9t+4 = 0 \\ \\ \Delta = 81 - 32 = 49 \\ \\ t_{1,2} = \dfrac{9\pm 7}{4} \Rightarrow \\ \\ t_1 = \dfrac{16}{4} =4>0\Rightarrow 2^x = 4 \Rightarrow x = 2 \\ \\ t_2 = \dfrac{2}{4} \Rightarrow t_2 = \dfrac{1}{2}> 0 \Rightarrow 2^x = \dfrac{1}{2} \Rightarrow x = -1\\ \\ \Rightarrow \boxed{S = \Big\{-1;~2\Big\}}$