Question 967249
<pre>
For the following sequences calculate the indicated term:
a) {{{a[n]=(-1)^n*(3n+2)}}}, {{{a[5]}}}

All you do is substitute 5 for n and simplify

{{{a[n]=(-1)^n*(3n^""+2)}}}
{{{a[5]=(-1)^5(3(5)^""+2)}}}
{{{a[5]=(-1)(15^""+2)}}}
{{{a[5]=(-1)(17)}}}
{{{a[5]=-17}}}

b) {{{a[n]=ln(e^n)}}}, {{{a[13]}}}

ln(e<sup>n</sup>) simplifies to just n, so the
problem is only

{{{a[n]=n}}}, {{{a[13]}}}

Just substitute 13 for n and have 

{{{a[13]=13}}} 





c) {{{a[1]=1}}}, {{{a[n+1]=2a[n]+1}}}, {{{a[4]}}} 

This is a recursive formula, not a general formula,
like the other two.  

1. You are given a<sub>1</sub>.
2. You use n=1, a<sub>1</sub> and the formula to find a<sub>2</sub>.
3. You use n=2, a<sub>2</sub> and the formula to find a<sub>3</sub>.
4. You use n=3, a<sub>3</sub> and the formula to find a<sub>4</sub>.

1. {{{a[1]=1}}}

2. Substitute 1 for n and 1 for a<sub>1</sub>, and
simplify:

{{{a[n+1]=2a[n]+1}}} 
{{{a[1+1]=2a[1]+1}}}
{{{a[2]=2(1)+1}}}
{{{a[2]=2+1}}}
{{{a[2]=3}}}

3. Substitute 2 for n and 3 for a<sub>2</sub>, and
simplify:

{{{a[n+1]=2a[n]+1}}}
{{{a[2+1]=2a[2]+1}}} 
{{{a[3]=2(3)+1}}}
{{{a[3]=6+1}}}
{{{a[3]=7}}}

4. Substitute 3 for n and 7 for a<sub>3</sub>, and 
simplify:

{{{a[n+1]=2a[n]+1}}}
{{{a[3+1]=2a[3]+1}}} 
{{{a[4]=2(7)+1}}}
{{{a[4]=14+1}}}
{{{a[4]=15}}}

Edwin</pre>