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

1 Answer
Aug 6, 2017

We aim to prove by Mathematical Induction that for #n in NN, n ge 1# then:

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

When #n=1# the given result gives:

# 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 #n=1#

Now, Let us assume that the given result is true when #n=m#, for some #m in NN, m gt 1#, in which case for this particular value of #m# we have:

# 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 #n=m+1#

So, we have shown that if the given result is true for #n=m#, then it is also true for #n=m+1#. But we initially showed that the given result was true for #n=1# and so it must also be true for #n=2, n=3, n=4, ... # and so on.

Hence, by the process of mathematical induction the given result is true for #n in NN, n ge 1# QED

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 #n+1# terms of a GP with:

First Term #a=x^0=1#,
Common Ratio #r=x#

And so, using the GP formula, the sum of the first #n# terms is given by:

# 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