# Prove by Mathematical Induction? : # sum_(k=0)^n x^k = (1-x^(n+1)) / (1-x) #

##### 1 Answer

We aim to prove by Mathematical Induction that for

# sum_(k=0)^n x^k = (1-x^(n+1)) / (1-x) #

When

# LHS = sum_(k=0)^n x^k = x^0 + x^1 = 1 +x #

# RHS = (1-x^2)/(1-x) = ( (1+x)(1-x) ) / (1-x) = 1 + x#

So the given result is **true** when

Now, Let us **assume** that the given result is true when

# sum_(k=0)^m x^k = (1-x^(m+1)) / (1-x) #

Adding the next term of the sum gives us:

# sum_(k=0)^(m+1) x^k = (sum_(k=0)^(m) x^k ) + x^(m+1) #

And using the above assumption, we have:

# sum_(k=0)^(m+1) x^k = (1-x^(m+1)) / (1-x) + x^(m+1) #

# " " = (1-x^(m+1) + (1-x)x^(m+1))/(1-x) #

# " " = (1-x^(m+1) + x^(m+1) + x x^(m+1))/(1-x) #

# " " = (1 + x x^(m+1))/(1-x) #

# " " = (1 + x^(m+2))/(1-x) #

# " " = (1 + x^((m+1)+1))/(1-x) #

Which is the given result with

So, we have shown that if the given result is true for

Hence, by the process of mathematical induction the given result is true for

**Supporting Note**

The above result should come as no surprise because:

# sum_(k=0)^n x^k = x^0+x^1+x^2 + ... + x^n #

Which are the

First Term

#a=x^0=1# ,

Common Ratio#r=x#

And so, using the GP formula, the sum of the first

# S_n = a(1-r^n)/(1-r) #

# \ \ \ \ = 1(1-x^(n+1))/(1-x) \ \ \ # (remember we have#n+1# terms)

# \ \ \ \ = (1-x^(n+1))/(1-x) # QED