determine t3,t5,and t10 for the sequences
a) tn=(2n-1)squared
b) tn=3n-1-2
c) tn= (n-1)/(2n-1)
Just substitute n=3, 5 and 10.
For example $\displaystyle t_{\color{red}5}=(2*{\color{red}5}-1)^2=(10-1)^2=81$
Are you sure it is this writing ? o.Ob) tn=3n-1-2
For example :c) tn= (n-1)/(2n-1)
$\displaystyle t_{\color{red}3}=\frac{{\color{red}3}-1}{2*{\color{red}3}-1}=\frac{2}{6-1}=\frac 25$
Is it clear ?
[quote=euclid2;186590]determine t3,t5,and t10 for the sequences
a) tn=(2n-1)squared
$\displaystyle {\text{General term, }}t_n = \left( {2n - 1} \right)^2 \hfill \\$
$\displaystyle {\text{Now, for }}t_3 {\text{ put }}n = 3, \hfill \\$
$\displaystyle
t_3 = \left( {2 \times 3 - 1} \right)^2 = 5^2 = 25 \hfill \\$
$\displaystyle {\text{Now, for }}t_5 {\text{ put }}n = 5, \hfill \\$
$\displaystyle t_5 = \left( {2 \times 5 - 1} \right)^2 = 9^2 = 81 \hfill \\$
$\displaystyle {\text{Now, for }}t_{10} {\text{ put }}n = 10, \hfill \\$
$\displaystyle t_{10} = \left( {2 \times 10 - 1} \right)^2 = 19^2 = 361 \hfill \\
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[quote=euclid2;186590]determine t3,t5,and t10 for the sequences
b) tn=3^{n-1} - 2
$\displaystyle {\text{General term, }}t_n = 3^{n - 1} - 2 \hfill \\$
$\displaystyle {\text{Now, for }}t_3 {\text{ put }}n = 3, \hfill \\$
$\displaystyle t_3 = 3^{3 - 1} - 2 = 3^2 - 2 = 9 - 2 = 7 \hfill \\$
$\displaystyle {\text{Now, for }}t_5 {\text{ put }}n = 5, \hfill \\$
$\displaystyle t_5 = 3^{5 - 1} - 2 = 3^4 - 2 = 81 - 2 = 79 \hfill \\$
$\displaystyle {\text{Now, for }}t_{10} {\text{ put }}n = 10, \hfill \\$
$\displaystyle t_{10} = 3^{10 - 1} - 2 = 3^9 - 2 = 19683 - 2 = 19681 \hfill \\
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