Question 1037988
<pre><b>
Make a Pascal's triangle down to the line that starts with 1 4

Start with this.

    1
   1 1
  1   1
 1     1
1       1

Add the two 1's on the second row and place it underneath
them on the third row:

    1
   1 1
  1 2 1
 1     1
1       1 

Add the 1 and the 2 on the 2nd row getting 3 
and place the 3 between them on the 3rd row

    1
   1 1
  1 2 1
 1 3   1
1       1 

Add the 2 and the 1 on the 3rd row getting 3 
and place the 3 between them on the 4th row 

    1
   1 1
  1 2 1
 1 3 3 1
1       1

Add the 1 and the 3 on the 4th row getting 4 
and place the 4 between them on the 5th row 

    1
   1 1
  1 2 1
 1 3 3 1
1 4     1

Add the 3 and the 3 on the 4th row getting 6 
and place the 6 between them on the 5th row 

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6   1

Add the 3 and the 1 on the 4th row getting 4 
and place the 4 between them on the 5th row 

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

I'll just do the last one:

3. (a + b)<sup>4</sup>

Take the bottom line of Pascal's triangle:

1     4     6     4     1
1a b +4a b +6a b +4a b +1a b   <-- put in "a, space, b"
1a<sup>4</sup>b +4a<sup>3</sup>b +6a<sup>2</sup>b +4a<sup>1</sup>b +1a b   <-- put in a's exponents <sup>4</sup>,<sup>3</sup>,<sup>2</sup>,<sup>1</sup>,<sup>0</sup>
1a<sup>4</sup>b<sup>0</sup>+4a<sup>3</sup>b<sup>1</sup>+6a<sup>2</sup>b<sup>2</sup>+4a<sup>1</sup>b<sup>3</sup>+1a<sup>0</sup>b<sup>4</sup>  <-- put in b's exponents <sup>0</sup>,<sup>1</sup>,<sup>2</sup>,<sup>3</sup>,<sup>4</sup> 

Now you can erase the 1's and b<sup>0</sup> and a<sup>0</sup>
because they are just 1's, too:

 a<sup>4</sup>  +4a<sup>3</sup>b +6a<sup>2</sup>b<sup>2</sup>+4a b<sup>3</sup>+   b<sup>4</sup> 

and close up the spaces:

a<sup>4</sup>+4a3b+6a<sup>2</sup>b<sup>2</sup>+4ab<sup>3</sup>+b<sup>4</sup>

Do the others the same way with the upper rows of 
Pascal's triangle.

Always use the row that has for its second number 
(after the 1) the exponent of (a+b).

Edwin</pre></b>