Question

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))`

Answer 1

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))`

Answer 2

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.!