Prove by Mathematical Induction? : # 1 + 2(1/2) + 3(1/2)^2 + 4(1/2)^3 + ...+ n(1/2)^(n-1) = 4 - ((n+2)/2^(n-1)) #

2 Answers
Aug 22, 2017

We seek to prove that:

# 1 + 2(1/2) + 3(1/2)^2 + 4(1/2)^3 + ...+ n(1/2)^(n-1) = 4 - ((n+2)/2^(n-1)) #

Or, alternatively, using Sigma Notation:

# sum_(r=0)^n r(1/2)^(r-1) = 4 - ((n+2)/2^(n-1)) #

So let us test this assertion using Mathematical Induction:

Induction Proof - Hypothesis

We aim to prove by Mathematical Induction that for #n in NN# that:

# sum_(r=0)^n r(1/2)^(r-1) = 4 - ((n+2)/2^(n-1)) # ..... [A]

Induction Proof - Base case:

We will show that the given result, [A], holds for and #n=1# (and actually for #n=0#)

When #n=0# the given result gives:

# LHS = 0 #
# RHS = 4 - ((2)/2^(-1)) = 0#

When #n=1# the given result gives:

# LHS = 1 #
# RHS = 4 - ((3)/2^(0)) = 1#

So the given result is true when #n=1# (and in fact #n=0#)

Induction Proof - General Case

Now, Let us assume that the given result [A] 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_(r=0)^m r(1/2)^(r-1) = 4 - ((m+2)/2^(m-1)) # ..... [B]

So, let us add the next term to the series, so that we have:

# sum_(r=0)^(m+1) r(1/2)^(r-1) = (sum_(r=0)^m r(1/2)^(r-1)) +(m+1)(1/2)^m #

Then using [B] we have:

# sum_(r=0)^(m+1) r(1/2)^(r-1) = 4 - ((m+2)/2^(m-1)) +(m+1)(1/2)^(m) #
# " " = 4 - ((m+2)/2^(m-1)) * 2/2 +(m+1)/(2^m) #
# " " = 4 - (2(m+2))/(2^m) +(m+1)/(2^m) #
# " " = 4 - (2(m+2) -(m+1))/(2^m) #
# " " = 4 - (2m+4 -m-1)/(2^m) #
# " " = 4 - (m+3)/(2^m) #
# " " = 4 - ((m+1)+2)/(2^(m+1)-1) #

Which is the given result [A] with #n=m+1#

Induction Proof - Summary

So, we have shown that if the given result [A] 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=0# and #n=1# 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 [A] is true for #n in NN# QED

Hence we have:

# sum_(r=0)^n r(1/2)^(r-1) = 4 - ((n+2)/2^(n-1)) #

Aug 25, 2017

Answer:

An Alternative Proof, without using the Induction is given in

the Explanation.

Explanation:

Respected Steve M. Sir has solved, as is required, the

Problem using the Principle of Mathematical Induction.

Let us, as an Aliter, prove it without it.

Let, #(1)...S_n=1+2(1/2)+3(1/2)^2+4(1/2)^3+...+(n-1)(1/2)^(n-2)+n(1/2)^(n-1).#

Multiplying both sides by, #1/2,# we get,

#(2)...1/2*S_n=1(1/2)+2(1/2)^2+3(1/2)^3+4(1/2)^4+...+(n-1)(1/2)^(n-1)+n(1/2)^n.#

Then, #(1)-(2) rArr S_n-1/2*S_n=1+(2-1)(1/2)+(3-2)(1/2)^2+(4-3)(1/2)^3+...+(n-(n-1))(1/2)^(n-1)-n(1/2)^n,#

#:. 1/2*S_n=1+{(1/2)+(1/2)^2+(1/2)^3+...+(1/2)^(n-1)}-n(1/2)^n.#

We easily recognise that the Series in #{......}# is Geometric,

having, #(n-1)" terms, the first term, "a=1/2="the common ratio "r;#

hence, its sum=#{a(1-r^(n-1))}/(1-r)#

#={cancel(1/2)(1-(1/2)^(n-1))}/cancel(1-(1/2))=1-(1/2)^(n-1).#

# :.1/2*S_n=1+{1-(1/2)^(n-1)}-n(1/2)^n, i.e., #

#=2-(1+1/2*n)(1/2)^(n-1), or, #

# rArr 1/2*S_n=2-1/2*((n+2)/2^(n-1))," giving, "#

# S_n=4-((n+2)/2^(n-1)),# as desired.

Enjoy Maths.!