For every n ∊ N, (1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2)) is greater than or equal to (1/4) +(1/(2^(n+1)))

 For every n ∊ N, 

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) is greater than or equal to (1/4) +(1/(2^(n+1)))

PROOF:

Base Case:

For n = 1,

(1- (1/2^(n))) = 1 - 1/2= 1/2

1/4 - (1/(2^(n+1))) = 1/4 +(1/2^(1+1)) = 1/4 +1/4 = 2/4 = 1/2

1/2 >= 1/2

Thus the summation holds for n = 1.

Induction Step:

Assume P(n) is true

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) >= (1/4) +(1/(2^(n+1)))

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n)))(1-(1/2^(n+1))) >= (1/4 + (1/2^(n+1)))(1/4 + (1/2^(n+1)))

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

= 1/4 + (1/2^(n+1))[1- (1/2^(n+1)) -1/4]

*since (1/2^(n+1)) + 1/4 < 1/2

--> 1 - ((1/2^(n+1)) - 1/4) > 1 - 1/2

--> 1 - ((1/2^(n+1)) - 1/4) > 1/2*

> 1/4 +(1/2^(n+1))(1/2) (from the * above)

= 1/4 + (1/2^(n+2)) = (1/2^((n+1)+1))

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n+1))) >= 1/4 + (1/2^((n+1)+1))

This proves that P(n+1) is true.

Thus by proof by mathematical induction,  For every n ∊ N, 

(1-(1/2))(1-(1/4))(1-(1/8))(1-(1/16))...(1-(1/2^(n))) is greater than or equal to (1/4) +(1/(2^(n+1)))

For every natural number 1/1 + 1/2 + 1/3 + ... + 1/((2^n)-1)+1/2^n \(\ge\) 1+ n/2

 For every natural number n,

 1/1 + 1/2 + 1/3 + ... + 1/((2^n)-1)+1/2^n \(\ge\) 1+ n/2

Base Case: n=1

1/(2^1-1)+1/2 \(\ge\) 1 + 1/2

3/2 \(\ge\) 3/2

Base case holds.

Induction Step:

Assume induction hypothesis in base case holds for n, then prove it holds for n+1 as well.

1/1 + 1/2 + 1/3 + ... + 1/((2^n+1)-1)+1/2^(n+1) \(\ge\) 1+ (n+1)/2

1 +  n/2 + 1/(2^(n+1)-1) + 1/2^(n+1) \(\ge\) 1 + (n+1)/2

1 + (n(2^(n+1)-1)/(2(2^(n+1)-1)) + 2/2(2^(n+1)-1) + 1/2^(n+1) \(\ge\) 1 + (n+1)/2

1 + (n((2^(n+1)-1)+2)/(2(2^(n+1)-1)) + 1/2^(n+1) \(\ge\) 1 + (n+1)/2

1+ (n+2)/2 + 1/2(n+1) \(\ge\) 1 + (n+1)/2

which is true, and proves our induction hypothesis.

For every \(n\in\mathbb{N}\), \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}

\[\sum_{k=1}^{n}\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!} Base Case -> (n=1) \frac{1}{(1+1)!}=1-\frac{1}{(1+1)!} \frac{1}{2}=1-\frac{1}{2} .5=.5 Induction Step -> Assume that for every \(n\in\mathbb{N}\) \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!} Since \frac{1}{2!}+...+\frac{n}{(n+1)!}+\frac{n+1}{((n+1)+1)!}=1-\frac{1}{((n+1)+1)!} Then \frac{1}{2!}+...+\frac{n}{(n+1)!}+\frac{n+1}{(n+2)!}=1-\frac{1}{(n+2)!}

For every n ∊ ℕ, 1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

For every n ∊ ℕ,  1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

Base case: Verify the statement for P(1) 

(1)(1+2) = (1)(1+1)(2(1)+7)/6

(1)(3) = (1)(2)(9)/6

3 = 18/6 → 3 = 3 ☑

Induction case: 

Suppose P(n) holds i.e  1·2 + 2·4 + 3·5 + . . . + n(n+2) = n(n+1)(2n+7)/6

Then P(n+1)

=  1·2 + 2·4 + 3·5 + . . . + (n+1)((n+1)+2) 

which simplifies to: 

= 1·2 + 2·4 + 3·5 + . . . + (n+1)(n+3) 

= (1·2 + 2·4 + 3·5 + . . . + n(n+2)) + (n+1)(n+3) = n(n+1)(2n+7)/6 + (n+1)(n+3)

P(n)  + (n+1)(n+3) = n(n+1)(2n+7)/6 + (n+1)(n+3)

the right hand side i.e.  n(n+1)(2n+7)/6 + (n+1)(n+3) can be simplified to 

= n(n+1)(2n+7)/6 + 6(n+1)(n+3)/6

factor out (n+1)/6 and get

= (n+1/6)(n(2n+7) + 6(n+3))

= (n+1/6)(2n^2 + 7n + 6n + 18)

=(n+1/6)((n+2)(n+9)

(n+1)(n+2)(n+9)/6 ☑

For every n ∊ N, 1・2 + 2・3 + 3・4 + ⋅ ⋅ ⋅ + n(n+1) = n(n+1)(n+2)/3

 Proof by Induction. 

The base case \((n=1):\) \(1*2=1(1+1)(1+2)/3.\) Which simplifies to \(2=2.\)

The induction step. 

The induction hypothesis: assume that for some \(n ∊ N.\) Then \(1*2 + 2*3 + 3*4 +⋅⋅⋅+ n(n+1)=n(n+1)(n+2)/3.\) We must show \(1・2+2・3+3・4+⋅⋅⋅+n(n+1)+(n+1)(n+2)=(n+1)(n+2)(n+3)/3.\) However, \(1*2 + 2*3 + 3*4 +⋅⋅⋅+ n(n+1)=n(n+1)(n+2)/3+(n+1)(n+2).\) \(n(n+1)(n+2)/3+(n+1)(n+2)\) can be rewritten as \(n/3(n+1)(n+2)+(n+1)(n+2).\) Then it can be factored out as \((n+1)(n+2))(n/3+1).\) Finally, it can be rewritten as \((n+1)(n+2)(n+3)/3.\) So, for every \(n ∊ N,\) \(1・2+2・3+3・4+⋅⋅⋅+n(n+1)=(n)(n+1)(n+2)/3.\)

Prove that for every natural number n, \(2^n+1 \leq 3^n \)

We will prove that for \(n \in \mathbb{N}, 2^n+1 \leq 3^n \) by induction. First we will set up a base case at \[ \begin{align} n&=1\\ 2^1+1&=3\\ 3^1&=3\\ so \ 2^1+1 &\leq 3^1 \end{align} \] Since this base case holds we will move on to the induction step and assume that for some \( n \in \mathbb{N}, \ 2^n+1 \leq 3^n \), and will show that this implies that \(2^{n+1}+1 \leq 3^{n+1}\) \[ \begin{align} 2^{n+1}+1 &= (2*2^n)+1\\ &= 2(2^n+1)-1\\ &=((2^n+1)+(2^n+1))-1\\ &\leq 3^n+3^n-1\\ &=3^{n+1}-1-(3^n)\\ &\leq3^{n+1} \end{align} \] So, by induction, for every \(n \in \mathbb{N}, 2^n+1 \leq 3^n\).

Prove that for every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)

 For every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)

Proof By Induction

Base Case: n = 1

\(1^{3}\) = 1

\(\frac{1^{2}(1+1)^{2}}{4}\) = 1

Base case holds!

Induction Hypothesis:

Assume that \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\) is true for some k \(\in\) \(\mathbb{N}\). We will use this statement in our inductive step. 

Inductive Step: 

Suppose P(k) is true, P(k): \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\). Prove then that P(k+1) is true. P(k+1): \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{(k+1)^{2}(k+2)^{2}}{4}\) is true for some k \(\in\) \(\mathbb{N}\) 

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\) + \((k+1)^{3}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2} + 4(k+1)^{3}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2} + 4(k+1)^{1}(k+1)^{2}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{(k+1)^{2}(k^{2} + 4k + 4)}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)

This result equals the right hand side of P(k+1). We have proven by induction that for every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)