Question 968173
a) 1/(1*2^2) + 1/(2*3^2) + 1/(3*4^2)
<pre>
Evaluating is just basic fraction math:

{{{1/(1*2^2) + 1/(2*3^2) + 1/(3*4^2)}}}
{{{1/(1*4) + 1/(2*9) + 1/(3*16)}}}
{{{1/4 + 1/18 + 1/48}}}
{{{36/144+8/144+3/144}}}
{{{47/144}}}

For the summation, I'll try some coloring to see if that helps:

{{{1/(red(1)*green(2)^2) + 1/(red(2)*green(3)^2) + 1/(red(3)*green(4)^2)}}}

Notice that the 1's on top stay the same.
Notice that the 2 exponents on the bottom stay the same in all terms.
Notice that the red numbers go <font color="red">1</font>,<font color="red">2</font>,<font color="red">3</font>
Notice that the green numbers go <font color="green">2</font>,<font color="green">3</font>,<font color="green">4</font>, just 1 more than the red numbers.
So if <font color="red">k</font> represents a red number, then the corresponding green number is <font color="green">k+1</font>.

So each term is of the form {{{1/(red(k)*green((k+1))^2)}}}

And <font color="red">k</font> starts at 1 and goes to 3.  We indicate that
with

{{{sum(1/(red(k)*green((k+1))^2),matrix(1,3,"","",k=1),matrix(1,5,"","","","",3))}}}
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</pre>
b) 3 + 6 + 9 + 12 + 15
<pre>
The evaluation is simple addition, getting 45.

This one is easier than the first one.

They are the first 5 multiples of 3.  They are

{{{3*1 + 3*2 + 3*3 + 3*4 + 3*5}}}

The {{{3*""}}} is the same on all the terms,
The numbers after the {{{3*""}}} go 1,2,3,4,5
So if k represents the numbers on the right of the {{{3*""}}},
then k starts at 1 and goes up to and including 5:

So each term has the form {{{3*k}}}

The summation is

{{{sum(3*k^"",k=1,matrix(1,3,"","",5))}}}

Edwin</pre>