# 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

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

# 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

When

# LHS = 0 #

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

When

# LHS = 1 #

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

So the given result is **true** when

**Induction Proof - General Case**

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

# 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

**Induction Proof - Summary**

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

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

Hence we have:

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

#### 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,

Multiplying both sides by,

Then,

We easily recognise that the **Series** in **Geometric,**

having,

hence, its **sum**=

**desired.**

**Enjoy Maths.!**