Question 936889
(1)Use the binomial theorem to find the 18th term in the binomial expansion of [2m-n(sqrt(2)]^27
<pre>
{{{(2m^""-n*sqrt(2))^27}}}

Usually the letter n is used to represent a general number, not a specific
number, but here n is used as a specific number, so I can't speak of the "nth
term", so I'll have to speak of the "pth" term instead to avoid conflict of
letters.

The pth term of {{{(q+r)^s}}}, if you start counting with 0 instead of 1, is

{{{(matrix(2,1,s,p))q^(s-p)*r^p}}}, 

where {{{(matrix(2,1,s,p))}}} is the binomial coefficient which is the same as
C(s,p) or sCp or "the number of combinations of s things taken p at a time."

The 18th term is the 17th term if we start counting from 0, so we
substitute  

{{{p=17}}}, {{{q=2m}}}, {{{r=-n*sqrt(2)}}}, {{{s=27}}}

into

{{{(matrix(2,1,s,p))q^(s-p)*r^p}}}

{{{(matrix(2,1,27,17))*(2m)^(27-17)*(-n*sqrt(2))^17}}} and simplify a little:

{{{(matrix(2,1,27,17))*(2m)^10*(-n)^17*(sqrt(2))^17}}}

Further simplifying the factors separately:
   {{{(matrix(2,1,27,17))}}} = {{{8436285}}} <-- using a TI-84 calculator
   {{{(2m)^10}}} = {{{2^10*m^10}}} = {{{1024m^10}}}
   {{{(-n)^17}}} = {{{-n^17}}}
   {{{(sqrt(2))^17}}} = {{{(sqrt(2))^(16+1)}}} = {{{(sqrt(2))^16*(sqrt(2))^1}}} = {{{(sqrt(2))^(2*8)*sqrt(2)}}} = {{{((sqrt(2)^"")^2)^8*sqrt(2)}}} = {{{2^8*sqrt(2)}}} = {{{256sqrt(2)}}}

Putting them all together:

{{{8436285*1024m^10*(-n^17)*256sqrt(2)}}}

{{{-2211521495040m^10n^17}}}  <-- answer

------------------------------------
</pre>
(2)Find the 69th number in the 72nd row (n=72) of Pascal’s triangle.
<pre>
Again we start counting from 0, so the 69th number starting counting with 1,
is the 68th number, when we start counting with 0 instead of 1. 

{{{(matrix(2,1,72,68))}}}{{{""=""}}}{{{"C(72,68)"}}}{{{""=""}}}{{{1028790}}}

If you can't use a calculator you can use 

{{{(matrix(2,1,72,68))}}}{{{""=""}}}{{{(matrix(2,1,72,72-68))}}}{{{""=""}}}{{{(matrix(2,1,72,4))}}}{{{""=""}}}{{{(72*71*70*69)/(4*3*2*1)}}}{{{""=""}}}{{{1028790}}}.

{{{1028790}}}  <-- answer

Edwin</pre>