Mathematical induction

[Johulus]

I can't solve this task and would appreciate some help and advice.

Task is:

\(\displaystyle 1\cdot 2^{n} + 2\cdot 2^{n-1} + 3\cdot 2^{n-2} + \dotsb + n\cdot 2 +(n+1)=2^{n+2} -(n+3) \)

I was taught that first I need to check if the first term of a series on the left is equal to the expression on the right for n=1. So I did.
But, it only works if I take first two terms from this series:

\(\displaystyle 1\cdot 2^{1} + 2\cdot 2^{0}=2^{3}-4 \quad 4=4 \)

After the basis of induction I proceeded to the induction hypothesis and I just did this:

\(\displaystyle n=k \\ 1\cdot 2^{k} + 2\cdot 2^{k-1} + 3\cdot 2^{k-2} + \dotsb + k\cdot 2 +(k+1)=2^{k+2} -(n+3) \)

Then I went to the inductive step. I rewrote everything on the left side and after that I should add the following term for n=k+1. So I did:

\(\displaystyle n=k+1 \\ 1\cdot 2^{k} + 2\cdot 2^{k-1} + 3\cdot 2^{k-2} + \dotsb + k\cdot 2 +(k+1) + 2\cdot (k+1) + (k+2)=2^{k+3} -(k+4) \\ 2^{k+2} -(n+3)+ 2\cdot (k+1) + (k+2)=2^{k+3} -(k+4) \)

But, there is something wrong with this because when I chose some integer for k and put it into this last expression the left and the right side aren't equal. Obviously there is some mistake, but I don't know what is it nor I know what I should have done. This task is so much confusing to me. I did all the other tasks by this manner and everything was fine. But what also confuses me here is: Should I put \(\displaystyle n=k+1 \) into just \(\displaystyle (n+1) \) in the inducive step or into \(\displaystyle n\cdot 2 \) and \(\displaystyle (n+1) \)?

For example, if I had:

\(\displaystyle 1+2+ \dotsb + n=\dfrac{n(n+1)}{2} \\ for\, n=1 \quad 1=\dfrac{1\cdot (1+1)}{2} \quad 1=1 \\ for\, n=k \quad 1+2+ \dotsb +k=\dfrac{k(k+1)}{2} \\ for\, n=k+1 \quad 1+2+ \dotsb + k +(k+1)=\dfrac{(k+1)(k+2)}{2} \\ \dfrac{k(k+1)}{2} +(k+1)=\dfrac{(k+1)(k+2)}{2} \\ \dfrac{(k+1)(k+2)}{2}=\dfrac{(k+1)(k+2)}{2} \)

What I did in my task above is the same, I believe it is, but at some point I made a mistake and I don't know what it is and how I should solve that task.

 

[ksdhart]

Then I went to the inductive step. I rewrote everything on the left side and after that I should add the following term for n=k+1. So I did:

\(\displaystyle n=k+1 \\ 1\cdot 2^{k} + 2\cdot 2^{k-1} + 3\cdot 2^{k-2} + \dotsb + k\cdot 2 +(k+1) + 2\cdot (k+1) + (k+2)=2^{k+3} -(k+4) \\ 2^{k+2} -(n+3)+ 2\cdot (k+1) + (k+2)=2^{k+3} -(k+4) \)

The problem I see is with this step right here. You appear to have plugged in n = (k+1), but only on the right side of the equation. Plugging in n = (k+1) to the left side gives you:

\(\displaystyle 1\cdot 2^{k+1} + 2\cdot 2^{(k+1)-1} + 3\cdot 2^{(k+1)-2} + \dotsb + (k+1)\cdot 2 +((k+1)+1)=2^{(k+1)+2} - ((k+1)+3) \)

Simplify that down and see where you can go with it.

 

[Ishuda]

I can't solve this task and would appreciate some help and advice.

Task is:

\(\displaystyle 1\cdot 2^{n} + 2\cdot 2^{n-1} + 3\cdot 2^{n-2} + \dotsb + n\cdot 2 +(n+1)=2^{n+2} -(n+3) \)

I was taught that first I need to check if the first term of a series on the left is equal to the expression on the right for n=1. So I did.
But, it only works if I take first two terms from this series:

\(\displaystyle 1\cdot 2^{1} + 2\cdot 2^{0}=2^{3}-4 \quad 4=4 \)

...

First of all that is not correct. The series starts with 1*2ⁿ and ends with n*2+n+1. For n=1, the beginning, 1*2ⁿ=1*2¹=2, is the same as the n*2=1*2=2 and the final term is n+1=1+1=2. Thus for n=1, the value is 2+2=4=2ⁿ⁺²-(n+3)=2³-4=8-4=4

Next, for convenience, let S_n designate the nth sum, i.e.
S_n= 1*2ⁿ + 2*2^n-1 + 3*2^n-2 + ... + (n-1) 2² + n*2 + n+1
Now consider S_n+1
S_n+1= 1*2ⁿ⁺¹ + 2*2ⁿ + 3*2^n-1 + ... + (n-1)*2³ + n 2² + (n+1)*2 + (n+1+1)
and the difference between the two
S_n+1-S_n= 1*(2ⁿ⁺¹-2ⁿ) + 2*(2ⁿ-2^n-1) + 3*(2^n-1-2^n-2) + ... + (n-1) (2³-2²)+ n*(2²-2) + (n+1)*(2-1) + (n+1+1)
= [1*2ⁿ + 2*2^n-1 + 3*2^n-2 + ... + (n-1) 2² + n*2 + n+1] + n+2
= ...

Thus, true for n=1. If true for n then ...

 

[Johulus]

\(\displaystyle S_{n}=1\cdot 2^{n} + 2\cdot 2^{n-1}+ 3\cdot 2^{n-2}+ \dotsb + n\cdot 2 + (n+1) \\ S_{n+1}= 1\cdot 2^{n+1} + 2\cdot 2^{n}+ 3\cdot 2^{n-1}+ \dotsb + (n+1)\cdot 2 + (n+2) \\ S_{n+1}-S_{n}=[1\cdot (2^{n+1}-2^{n}) + 2\cdot (2^{n}-2^{n-1}) + 3\cdot (2^{n-1}-2^{n-2}) + \dotsb +n(2^{2}-2)+(n+1)(2-1)] +(n+2) \\ S_{n+1}-S_{n}=S_{n} + (n+2) \\ S_{n+1}=2\cdot S_{n} + (n+2) \\ S_{n}=2^{n+2}-(n+3) \\ 2\cdot [2^{n+2}-(n+3)]+ (n+2)=2^{n+3}-(n+4) \)

Hope this is good.

First of all that is not correct. The series starts with 1*2ⁿ and ends with n*2+n+1. For n=1, the beginning, 1*2ⁿ=1*2¹=2, is the same as the n*2=1*2=2 and the final term is n+1=1+1=2. Thus for n=1, the value is 2+2=4=2ⁿ⁺²-(n+3)=2³-4=8-4=4

But, I don't get it what this means. So what if \(\displaystyle 1\cdot 2^{n}=n\cdot 2 = 2\) and if \(\displaystyle n+1=2 \) for \(\displaystyle n=1 \)? What does that have to do with \(\displaystyle 2+2=4=2^{n+2}-(n+3)=2^{3}-4=4 \) ? Where from did you get that 2+2?

 

[Deleted member 4993]

Any reason why you're not showing this way:
\(\displaystyle 1\cdot 2^{n} + 2\cdot 2^{n-1} + 3\cdot 2^{n-2} + \dotsb + n\cdot 2 +2n+4=2^{n+2} \)

Because that would be wrong - instead of '+4' that should be '-2'.

6 minutes in the corner......

 

[Ishuda]

\(\displaystyle S_{n}=1\cdot 2^{n} + 2\cdot 2^{n-1}+ 3\cdot 2^{n-2}+ \dotsb + n\cdot 2 + (n+1) \\ S_{n+1}= 1\cdot 2^{n+1} + 2\cdot 2^{n}+ 3\cdot 2^{n-1}+ \dotsb + (n+1)\cdot 2 + (n+2) \\ S_{n+1}-S_{n}=[1\cdot (2^{n+1}-2^{n}) + 2\cdot (2^{n}-2^{n-1}) + 3\cdot (2^{n-1}-2^{n-2}) + \dotsb +n(2^{2}-2)+(n+1)(2-1)] +(n+2) \\ S_{n+1}-S_{n}=S_{n} + (n+2) \\ S_{n+1}=2\cdot S_{n} + (n+2) \\ S_{n}=2^{n+2}-(n+3) \\ 2\cdot [2^{n+2}-(n+3)]+ (n+2)=2^{n+3}-(n+4) \)

Hope this is good.

That show 'if n then n+1' step, You need the words to go with all of it and the demonstration for n=1.

But, I don't get it what this means. So what if \(\displaystyle 1\cdot 2^{n}=n\cdot 2 = 2\) and if \(\displaystyle n+1=2 \) for \(\displaystyle n=1 \)? What does that have to do with \(\displaystyle 2+2=4=2^{n+2}-(n+3)=2^{3}-4=4 \) ? Where from did you get that 2+2?

In a different way, there are always at least two terms in the series, the second to last n*2 and final n+1. If the first term 1*2ⁿ is the same as the second to last term, then there are only the two terms. For n=1, the first term is the same as the second to last term so there are two terms 1*2=2 and n+1=2. The sum of these is 4 so S₁=4