Question 824889
find the tenth term in the expansion of (x+3)^12
<pre>
The {{{R}}}th term of {{{(A+B)^N}}} is given by
this expression:

{{{(matrix(2,1,N,R-1))A^(N-R+1)B^(R-1)}}}

where {{{(matrix(2,1,p,q))}}} means the same as {{{C(p,q)}}} or {{{pCq}}} 

In your problem {{{A=x}}}, {{{B=3}}}, {{{N=12}}}, {{{R=10}}}

So we substitute those and get

 {{{(matrix(2,1,12,10-1))(x)^(12-10+1)(3)^(10-1)}}}

{{{(matrix(2,1,12,9))(x)^(3)(3)^(9)}}}

{{{(220)(x^3)(19683)}}}

{{{4330260x^3}}}

The entire expansion is:

{{{x^12+36x^11+594x^10+5940x^9+40095x^8+192456x^7+673596x^6+1732104x^5+3247695x^4+4330260x^3+3897234x^2+2125764x+531441}}}

Edwin</pre>