Mathematical induction problem

[DDarthVader]

Hello! First of all I have like 5 exercises I don't quite understand so will it be a problem if I create 5 new topics in the next 24h?

Homework Statement

Prove, by using mathematical induction that if $x+1 \geq 0$ then $(1+x)^n \geq 1+nx$.

Homework Equations

The Attempt at a Solution

Basic step: If $n=1$ then $1+x \geq 1+x$ which is true.
Induction Step: Now making $n=k$ we get $(1+x)^k \geq 1+kx$. If the hypothesis holds for $n=k$ then it will hold for $n=k+1$. Making $n=k+1$ we get $(1+x)^k + (1+x)^{k+1} \geq 1+(k+1)x$. And from this:
$(1+x)^k(1+(1+x)) \geq 1+(k+1)x$
But by our induction hypothesis $(1+x)^k \geq 1+kx$ which means that $(1+x)^k(1+(1+x)) \geq 1+(k+1)x$ is true.

Thanks!

 

[SammyS]

Hello! First of all I have like 5 exercises I don't quite understand so will it be a problem if I create 5 new topics in the next 24h?

Homework Statement

Prove, by using mathematical induction that if $x+1 \geq 0$ then $(1+x)^n \geq 1+nx$.

Homework Equations

The Attempt at a Solution

Basic step: If $n=1$ then $1+x \geq 1+x$ which is true.
Induction Step: Now making $n=k$ we get $(1+x)^k \geq 1+kx$. If the hypothesis holds for $n=k$ then it will hold for $n=k+1$. Making $n=k+1$ we get
This looks like a typo. → $(1+x)^k + (1+x)^{k+1} \geq 1+(k+1)x$.
And from this:
$(1+x)^k(1+(1+x)) \geq 1+(k+1)x$
But by our induction hypothesis $(1+x)^k \geq 1+kx$ which means that $(1+x)^k(1+(1+x)) \geq 1+(k+1)x$ is true.

Thanks!

The induction step is:

Assume the following is true: $(1+x)^k \geq 1+kx\,.$

From that you need to show that the following is true: $(1+x)^{k+1} \geq 1+(k+1)x\,.$​

It looks to me as if you're assuming the hypothesis holds for n = k+1 .

 

[DDarthVader]

The induction step is:

Assume the following is true: $(1+x)^k \geq 1+kx\,.$

From that you need to show that the following is true: $(1+x)^{k+1} \geq 1+(k+1)x\,.$​

It looks to me as if you're assuming the hypothesis holds for n = k+1 .

That typo is actually not a typo. And I'm trying to say that if the $n=k$ holds then I'll try to prove that $n=k+1$ also holds by doing $(1+x)^k + (1^x)^{k+1}$. I can write it clearer in my language. But the main problem here is $(1+x)^k + (1+x)^{k+1}$. Is this correct?

 

[SammyS]

That typo is actually not a typo. And I'm trying to say that if the $n=k$ holds then I'll try to prove that $n=k+1$ also holds by doing $(1+x)^k + (1^x)^{k+1}$. I can write it clearer in my language.

No, the following is not correct.

But the main problem here is $(1+x)^k + (1+x)^{k+1}$. Is this correct?

If you substitute k+1 for n in $\displaystyle (1+x)^n \geq 1+nx\,,$ then you get $\displaystyle (1+x)^{k+1} \geq 1+(k+1)x\,.$

By the Way: $\displaystyle (1+x)^{k+1}=(1+x)^k\cdot(1+x)\,,$ it's not the same as $(1+x)^k+(1+x)^{k+1}\,.$

Maybe you're thinking of $\displaystyle \sum_{n=0}^{k+1}(1+x)^n=\left(\sum_{n=0}^{k}(1+x)^n\right)+(1+x)^{k+1}\,.$