Respuesta :
Answer:
[tex]\textsf{a)} \quad x = 17[/tex]
[tex]\textsf{b)} \quad T_n=3n^2-3n-1[/tex]
[tex]\textsf{a)} \quad x = 20[/tex]
[tex]\textsf{b)} \quad T_n=3n^2-4n+5[/tex]
Step-by-step explanation:
Given quadratic number pattern:
- -1, 5, x, 35, ...
To find the equation for the nth term, we can use the general form of a quadratic equation:
[tex]\boxed{T_n=an^2 + bn + c}[/tex]
where n is the position of the term.
Let's substitute the values of T₁, T₂ and T₄ into the quadratic equation: to create three equations:
[tex]\begin{aligned}T_1=a(1)^2+b(1)+c&=-1\\a+b+c&=-1\end{aligned}[/tex]
[tex]\begin{aligned}T_2=a(2)^2+b(2)+c&=5\\4a+2b+c&=5\end{aligned}[/tex]
[tex]\begin{aligned}T_4=a(4)^2+b(4)+c&=35\\16a+4b+c&=35\end{aligned}[/tex]
Rearrange the first equation to isolate c:
[tex]c=-a-b-1[/tex]
Substitute this into the second and third equations:
[tex]\begin{aligned}4a+2b+(-a-b-1)&=5\\3a+b&=6\end{ailgned}[/tex]
[tex]\begin{aligned}16a+4b+(-a-b-1)&=35\\15a+3b&=36\end{ailgned}[/tex]
Solve the equations simultaneously by rearranged the first equation to isolate b and substituting this into the second equation and solving for a:
[tex]b=-3a+6[/tex]
[tex]\begin{aligned}15a+3(-3a+6)&=36 \\15a-9a+18&=36\\6a&=18\\a&=3 \end{aligned}[/tex]
Substitute the found value of a into the equation for b and solve for b:
[tex]\begin{aligned}b&=-3a+6\\&=-3(3)+6\\&=-9+6\\&=-3\end{aligned}[/tex]
Finally, substitute the found values of a and b into the equation for c and solve for c:
[tex]\begin{aligned}c&=-a-b-1\\&=-3-(-3)-1\\&=-3+3-1\\&=-1\end{aligned}[/tex]
Therefore, the equation for the nth term is:
[tex]\boxed{T_n=3n^2-3n-1}[/tex]
The value of x is the 3rd term. Therefore, to find the value of x, substitute n = 3 into the equation for the nth term:
[tex]\begin{aligned}T_3&=3(3)^2-3(3)-1\\&=3(9)-3(3)-1\\&=27-9-1\\&=18-1\\&=17\end{aligned}[/tex]
Therefore, the value of x is 17.
[tex]\hrulefill[/tex]
Given quadratic number pattern:
- 4, 9, x, 37, ...
To find the equation for the nth term, we can use the general form of a quadratic equation:
[tex]\boxed{T_n=an^2 + bn + c}[/tex]
where n is the position of the term.
Let's substitute the values of T₁, T₂ and T₄ into the quadratic equation: to create three equations:
[tex]\begin{aligned}T_1=a(1)^2+b(1)+c&=4\\a+b+c&=4\end{aligned}[/tex]
[tex]\begin{aligned}T_2=a(2)^2+b(2)+c&=9\\4a+2b+c&=9\end{aligned}[/tex]
[tex]\begin{aligned}T_4=a(4)^2+b(4)+c&=37\\16a+4b+c&=37\end{aligned}[/tex]
Rearrange the first equation to isolate c:
[tex]c=-a-b+4[/tex]
Substitute this into the second and third equations:
[tex]\begin{aligned}4a+2b+(-a-b+4)&=9\\3a+b&=5\end{ailgned}[/tex]
[tex]\begin{aligned}16a+4b+(-a-b+4)&=37\\15a+3b&=33\end{ailgned}[/tex]
Solve the equations simultaneously by rearranged the first equation to isolate b and substituting this into the second equation and solving for a:
[tex]b=-3a+5[/tex]
[tex]\begin{aligned}15a+3(-3a+5)&=33 \\15a-9a+15&=33\\6a&=18\\a&=3 \end{aligned}[/tex]
Substitute the found value of a into the equation for b and solve for b:
[tex]\begin{aligned}b&=-3a+5\\&=-3(3)+5\\&=-9+5\\&=-4\end{aligned}[/tex]
Finally, substitute the found values of a and b into the equation for c and solve for c:
[tex]\begin{aligned}c&=-a-b+4\\&=-3-(-4)+4\\&=-3+4+4\\&=5\end{aligned}[/tex]
Therefore, the equation for the nth term is:
[tex]\boxed{T_n=3n^2-4n+5}[/tex]
The value of x is the 3rd term. Therefore, to find the value of x, substitute n = 3 into the equation for the nth term:
[tex]\begin{aligned}T_3&=3(3)^2-4(3)+5\\&=3(9)-4(3)+5\\&=27-12+5\\&=15+5\\&=20\end{aligned}[/tex]
Therefore, the value of x is 20.
