We are asked to evaluate each of the expressions for n = 1, 2, 3, 4
We simply need to substitute the value of n into the expression and simplify the expression.
Expression 1:
[tex]\begin{gathered} for\; n=1\colon\; \; 4n-1=4(1)-1=4-1=3 \\ for\; n=2\colon\; \; 4n-1=4(2)-1=8-1=7 \\ for\; n=3\colon\; \; 4n-1=4(3)-1=12-1=11 \\ for\; n=4\colon\; \; 4n-1=4(3)-1=16-1=15 \end{gathered}[/tex]
Expression 2:
[tex]\begin{gathered} for\; n=1\colon\; \; 3-n^2=3-(1)^2=3-1=2 \\ for\; n=2\colon\; \; 3-n^2=3-(2)^2=3-4=-1 \\ for\; n=3\colon\; \; 3-n^2=3-(3)^2=3-9=-6 \\ for\; n=4\colon\; \; 3-n^2=3-(4)^2=3-16=-13 \end{gathered}[/tex]
Expression 3:
[tex]\begin{gathered} for\; n=1\colon\; \; \frac{1}{n-2}=\frac{1}{1-2}=\frac{1}{-1}=-1 \\ for\; n=2\colon\; \; \frac{1}{n-2}=\frac{1}{2-2}=\frac{1}{0}=\text{undefined} \\ for\; n=3\colon\; \; \frac{1}{n-2}=\frac{1}{3-2}=\frac{1}{1}=1 \\ for\; n=4\colon\; \; \frac{1}{n-2}=\frac{1}{4-2}=\frac{1}{2}=0.5 \end{gathered}[/tex]
Expression 4:
[tex]\begin{gathered} for\; n=1\colon\; \; \frac{n^2}{n-1}=\frac{(1)^2}{1-1}=\frac{1}{0}=\text{undefined} \\ for\; n=2\colon\; \; \frac{n^2}{n-1}=\frac{(2)^2}{2-1}=\frac{4}{1}=4 \\ for\; n=3\colon\; \; \frac{n^2}{n-1}=\frac{(3)^2}{3-1}=\frac{9}{2}=4.5 \\ for\; n=4\colon\; \; \frac{n^2}{n-1}=\frac{(4)^2}{4-1}=\frac{16}{3}=5.3 \end{gathered}[/tex]
Expression 5:
[tex]\begin{gathered} for\; n=1\colon\; \; 2n+4=2(1)+4=2+4=6 \\ for\; n=2\colon\; \; 2n+4=2(2)+4=4+4=8 \\ for\; n=3\colon\; \; 2n+4=2(3)+4=6+4=10 \\ for\; n=4\colon\; \; 2n+4=2(4)+4=8+4=12 \end{gathered}[/tex]
