Question 874363
<pre>
The binomial theorem is

{{{(x+y)^n}}}{{{""=""}}}{{{sum(("C(n,r)"x^(n-r)y^r),r=0,matrix(1,3,"","",n))}}}
</pre>
1.Prove that 3^n=Summation(r=0,n)2^rC(n,r)
<pre>
Since 3 = 1+2, let x=1 and y=2 in the binomial theorem

{{{3^n}}}{{{""=""}}}{{{(1+2)^n}}}{{{""=""}}}{{{sum(("C(n,r)"1^(n-r)2^r),r=0,matrix(1,3,"","",n))}}}{{{""=""}}}{{{sum( (2^r*"C(n,r)") ,r=0,matrix(1,3,"","",n)))}}}

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2. Evaluate  summation(n=1,20)(1.1)^n
<pre>
{{{sum(1.1^n,n=1,matrix(1,3,"","",20))}}}{{{""=""}}}{{{1.1^1+1.1^2+""*""*""*""+1.1^20}}}

That's the sum of a geometric series with a<sub>1</sub> = 1.1, r=1.1, n=20

Substitute in the sum formula for a geometric series:

{{{S[n]}}}{{{""=""}}}{{{(a[1](r^n-1))/(r-1)}}}

{{{S[20]}}}{{{""=""}}}{{{(1.1(1.1^20-1))/(1.1-1)}}}{{{""=""}}}{{{(1.1(1.1^20-1))/0.1}}}{{{""=""}}}{{{(1.1(6.7245-1))/0.1}}}{{{""=""}}}{{{(1.1(5.7245))/0.1}}}{{{""=""}}}{{{expr(1.1/0.1)*(5.7245)}}}{{{""=""}}}{{{11(5.7245)}}}{{{""=""}}}{{{63.0025}}}

Edwin</pre>