Resolução testesdevestibulares

1/ (10 0)(10 1)(10 2)...(10 10)\nigual linha 10 do triângulo do Pascal\n1 10 95 20 20 252 20 220 45 10 1\nlogo: 1-10-45-120-210-252-210-220-45-10 + 1 = 0 alternativa e\n\n2) a2n 32768 = 2^n 2^15\n10=15 alternativa a\n\n3) O termo médio é o 6 ° termo. logo p=5.\n\nT_p+1 = (n p)(x ^p)\n8064x^5 = (10 9)(x ^1)(ky)^5\nK^5 = 8064x^5/ 258048x^5 = K^5 / 1/32\nK=1/2\n\n4) T_p+1=(6) (2/ 3)^n(1 - 1/3)^p\nT_p+1 = (6)(p)(3^(-p))\nno termo independente termos 12-3p=12-30= p=4\n\n5) No 4° temos T_p+1 = 4\n\nT_4 = (n 3) 3 ^3\nNo 8° temos T_p+1 = 8\n\nT_8 = (n 7)(x -7)\nse (n 3) = (n 7) temos p=7-n\n3t=n\n\nlogo n=10 alternativa\nc 6) T_p+1 = (7 p)(K^2)^(7 -p)(1 / x^ 5)\n\ntP+1 = (p)(K^7-p)\nT_p+1 = (7)(K^7 - 2p) / x ^ 5\nno termo independente termos 14 - 7 = 0\np=2\n\n7) (1 + x)^(n) = (1 + α)^(n)\n= 3^n alternativa b\n\nb) T_p= (6)(n - p)( x ^p)\n(6)(p) \nno termo independent termos 6 - 3p = p=2\n\nT_3 = (2 2) ^2 == 3^2 (1/ 4)\n\nT_3 = (6)(1)(1) ^2 = 6.1.1 / 4 = 6! = 6 = 0.75\n\nalternativa a\n\n9) No 6 temos T_p1 = 6\np=5\n\nTS = (1)(1)(0.5)^5 = 2^5 = 2^5.5 = am (ab) = (a + b)^10 = (ab)^(1)\ncomo termos (ab = 2) temos L = 1 + 2^2 (10 1) = 1024\n\nalternativa e