Question 968052
<pre>
{{{b[1] = 8}}}, {{{b[k + 1] = 2b[k] + 1}}}, {{{k >= 1}}}

The first term is given as {{{b[1]=8}}}

To find b<sub>2</sub>, plug in k=1 and {{{b[1]=8}}} and simplify:

{{{b[k+1] = 2b[k] + 1}}}
{{{b[1 + 1] = 2b[1] + 1}}}
{{{b[2] = 2(8) + 1}}}
{{{b[2] = 16 + 1}}}
{{{b[2] = 17}}}

To find b<sub>3</sub>, plug in k=2 and {{{b[2]=17}}} and simplify:

{{{b[k + 1] = 2b[k] + 1}}}
{{{b[2 + 1] = 2b[2] + 1}}}
{{{b[3] = 2(17) + 1}}}
{{{b[3] = 34 + 1}}}
{{{b[3] = 35}}}

To find b<sub>4</sub>, plug in k=3 and {{{b[3]=35}}} and simplify:

You finish.

Edwin</pre>