SOLUTION: 1.Prove that 3^n=Summation(r=0,n)2^rC(n,r) 2. Evaluate summation(n=1,20)(1.1)^n

Algebra ->  Permutations -> SOLUTION: 1.Prove that 3^n=Summation(r=0,n)2^rC(n,r) 2. Evaluate summation(n=1,20)(1.1)^n      Log On


   

Question 874363: 1.Prove that 3^n=Summation(r=0,n)2^rC(n,r)
2. Evaluate summation(n=1,20)(1.1)^n
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The binomial theorem is

%28x%2By%29%5En%22%22=%22%22

1.Prove that 3^n=Summation(r=0,n)2^rC(n,r)
Since 3 = 1+2, let x=1 and y=2 in the binomial theorem

3%5En%22%22=%22%22%281%2B2%29%5En%22%22=%22%22%22%22=%22%22

2. Evaluate summation(n=1,20)(1.1)^n
sum%281.1%5En%2Cn=1%2Cmatrix%281%2C3%2C%22%22%2C%22%22%2C20%29%29%22%22=%22%221.1%5E1%2B1.1%5E2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2B1.1%5E20

That's the sum of a geometric series with a1 = 1.1, r=1.1, n=20

Substitute in the sum formula for a geometric series:

S%5Bn%5D%22%22=%22%22%28a%5B1%5D%28r%5En-1%29%29%2F%28r-1%29

S%5B20%5D%22%22=%22%22%281.1%281.1%5E20-1%29%29%2F%281.1-1%29%22%22=%22%22%281.1%281.1%5E20-1%29%29%2F0.1%22%22=%22%22%281.1%286.7245-1%29%29%2F0.1%22%22=%22%22%281.1%285.7245%29%29%2F0.1%22%22=%22%22expr%281.1%2F0.1%29%2A%285.7245%29%22%22=%22%2211%285.7245%29%22%22=%22%2263.0025

Edwin