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mikky05v

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I am having issues with my homework questions, I can get them all the way to the very alst step but i seem to consistently get stuck trying to connect them to the answer at the end. (this will make more sense when you see the problem) I'm going to post both proofs I've worked on so far one after the other.

[STRIKE]Problem 1

Prove by induction: For all n that are an element of the natural numbers, Ʃ (i=1, n) 4

Proof :

1. For n=1, the LHS(left hand side)= 4

This TCIT for n=1

2. Assume TCIT for n=k, so Ʃ(i=1, k) 4i = (4

for n=k+1, LHS is Ʃ(i=1, k) 4i = (4

= Ʃ(i=1, k) 4i + 4

= (4

and this is where I get stuck, I know i need to get it to work out to be (4

[/STRIKE]

Problem 2

Prove by induction: For all n that are an element of the natural numbers, 1*1!+2*2!+...+n*n! = (n+1)!-1

Proof:

1. for n=1, LHS=1*1! = 1 while RHS = (1+1)! -1 = 1

thus TCIT for n=1

2. Assume TCIT for n=k, so 1*1!+2*2!+...+k*k! = (k+1)! -1

for n=k+1 the LHS = (1*1!+2*2!+...+k*k!)+(k+1)(k+1)!

=(k+1)!-1 + (k+1)(K+1)! by assumption

and again I can't figure out how to get it to come out to ((k+1)+1)!-1

[STRIKE]Problem 1

## Homework Statement

Prove by induction: For all n that are an element of the natural numbers, Ʃ (i=1, n) 4

^{i}= (4^{n+1}-4)/3## Homework Equations

## The Attempt at a Solution

Proof :

1. For n=1, the LHS(left hand side)= 4

^{1}while the RHS(right hand side)= (4^{1+1}-4)/3 = 12/3 = 4This TCIT for n=1

2. Assume TCIT for n=k, so Ʃ(i=1, k) 4i = (4

^{k+1}-4)/3for n=k+1, LHS is Ʃ(i=1, k) 4i = (4

^{1}+4^{2}+...+4^{k})+(4^{k+1})= Ʃ(i=1, k) 4i + 4

^{k+1}= (4

^{k+1}-4)/3 + 4^{k+1}by assumptionand this is where I get stuck, I know i need to get it to work out to be (4

^{(k+1)+1}-4)/3 but I don't see how[/STRIKE]

Problem 2

## Homework Statement

Prove by induction: For all n that are an element of the natural numbers, 1*1!+2*2!+...+n*n! = (n+1)!-1

## The Attempt at a Solution

Proof:

1. for n=1, LHS=1*1! = 1 while RHS = (1+1)! -1 = 1

thus TCIT for n=1

2. Assume TCIT for n=k, so 1*1!+2*2!+...+k*k! = (k+1)! -1

for n=k+1 the LHS = (1*1!+2*2!+...+k*k!)+(k+1)(k+1)!

=(k+1)!-1 + (k+1)(K+1)! by assumption

and again I can't figure out how to get it to come out to ((k+1)+1)!-1

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