Question

determinati n si x din dezvoltarea ( 3^{x/2} +3^{1-x/2} )^n stiind ca suma coeficientilor binomiali ai primilor trei termeni este egala cu 22iar suma dintre termenul al treilea si al cincilea este 420

Answer 1

$\text{Mai intai il aflam pe n:}\\ C_n^0+C_n^1+C_n^2= 22\\ 1+n+\dfrac{n(n-1)}{2}=22 |\cdot 2\\ 2+2n+n^2-n=44\\ n^2+n-42=0\\ \Delta= 1+4\cdot 42= 1+168=169\Rightarrow \sqrt{\Delta}=13\\ n_1=\dfrac{-1+13}{2}=6\\ n_2=\dfrac{-1-13}{2}=-7 \notin \mathbb{N},\text{deci nu convine}$

$\text{Acum il putem afla pe x:}\\ T_3=C_6^2 \left(3^{\frac{x}{2}}\right)^2 \cdot \left(3^{\frac{1-x}{2}}\right)^4 =\dfrac{6!}{4!\cdot 2!}\cdot 3^x \cdot 3^{2(1-x)} =15 \cdot 3^{2-x}\\ T_5= C_6^4 \left(3^{\frac{x}{2}}\right)^4 \cdot \left(3^{\frac{1-x}{2}}\right)^2 = \dfrac{6!}{4!\cdot 2!}\cdot 3^{2x}\cdot 3^{1-x} =15\cdot 3^{1+x}\\ T_3+T_5=420\\ 15(3^{2-x}+3^{1+x})=420\\ \dfrac{3^2}{3^x}+3\cdot 3^x =28\\ \dfrac{9}{3^x}+3\cdot 3^x =28 \\ 3^x \stackrel{not}{=} t>0\\ \dfrac{9}{t}+3t=28\\ 9+3t^2=28t$

$3t^2-28t+9=0\\ \Delta=784- 4\cdot 9\cdot 3=784- 108=676\Rightarrow \sqrt{\Delta}=26\\ t_1= \dfrac{28+26}{6}=\dfrac{54}{6}=9\\ t_2= \dfrac{28-26}{6}=\dfrac{2}{6}=\dfrac{1}{3}\\ i)\text{Pentru }t=9\Rightarrow x=2\\ ii)\text{Pentru }t=\dfrac{1}{3}\Rightarrow x=-1\\ S:x\in \{2,-1\}$