SOLUTION: Prove using simple induction that, for each integer n >= 1 3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2

Algebra ->  Proofs -> SOLUTION: Prove using simple induction that, for each integer n >= 1 3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2       Log On


   

Question 736005: Prove using simple induction that, for each integer n >= 1
3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Prove using simple induction that, for each integer n >= 1
3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2
----
Show it is true for n = 1
3 = (3^(1+1)-3)/2 = (3^2 - 3)/2 = 6/2 = 3
----------------------------------------
Assume it is true for n = k:
3 + 3^2 +...+3^k = (3^(k+1) -3)/2
----
Show it must be true for n = k+1
[3 + 3^2+...+3^k] + 3^(k+1)
---
= (3^(k+1) -3)/2 + 3^(k+1)
-----
= [3^(k+1) - 3) + 2*3^(k+1)]/2
----
= [3*3^(k+1) - 3]/2
-----
= [3^((k+1)+1) - 3]/2
===========================
Cheers,
Stan H.
==============================