Question 1082979
<pre><b>
{{{a[n]= (-2)^n/(n!)}}}

Let's plug in 1 for n and see if we get -2 for the 1st term.

{{{a[1]= (-2)^1/(1!)}}}
{{{a[1]= (-2)/1}}}
{{{a[1]= -2}}}

So the 1st term is -2.  So far so good.

Now, let's plug in 2 for n and see if we get 2 for the 2nd term.

{{{a[2]= (-2)^2/(2!)}}}
{{{a[1]= 4/(2*1)}}}
{{{a[1]= 4/2}}}
{{{a[2]=2}}}

So the 2nd term is 2.  So far so good.

Now, let's plug in 3 for n and see if we get -4/3 for the 3rd term

{{{a[3]= (-2)^3/(3!)}}}
{{{a[3]= (-8)/(3*2*1)}}}
{{{a[3]= (-8)/6}}}
{{{a[3]= -8/6}}}
{{{a[3]=-4/3}}}

So the 3rd term is -4/3.  So far so good.

Finally, let's plug in 4 for n and see if we get 2/3 for the 4th term.

{{{a[3]= (-2)^4/(4!)}}}
{{{a[3]= 16/(4*3*2*1)}}}
{{{a[3]= 16/24}}}
{{{a[3]= 2/3}}}

Yep!  So that proves that the answer is "TRUE".

Edwin</pre>