Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I get the number of different degrees of $n$ that would generate from $(n + 1)^k$.

For example, for $k = 2$, I want to get $3$ -- ($n^2+2n +1$).

Thanks

share|improve this question
1  
Do you mean the number of terms in the expansion of $(n+1)^k$? –  Andres Caicedo May 5 '12 at 21:34
5  
You would get $k+1$ distinct powers of $x$, if I am understanding your question right. See en.wikipedia.org/wiki/Binomial_theorem –  Galois Group May 5 '12 at 21:34

2 Answers 2

up vote 1 down vote accepted

the proof:

For $k=2$ we have the number of different degrees of $n$ is $3$ (because $(n+1)^2=n^2+2n+1$), we suppose that the number of different degrees of $n$ is $k+1$ for $(n+1)^k$,

so we have:

$$(n+1)^{k+1}=(n+1)^{k+1}(n+1)=n(n+1)^{k+1}+(n+1)^{k+1}$$

by hypothesis we know that the number of different degrees of $n$ of $(n+1)^k$ is $k+1$, multiplying by $1$ we get the same elements of the same degrees and by multiplying by $n$ we get another different degree which does not exists is the degree $k+1$ (of $n^{k+1}$), so we conclude that the number of different degrees of $(n+1)^{k+1}$ is $k+2$.

Finally we conclude by induction principle that the number of different degrees of $(n+1)^k$ is $k+1$.

share|improve this answer
    
Perhaps some discussion of why none of the powers of $n^j$ in $(n+1)^k$ is missing for $0\le j\le k$ would be useful. –  robjohn May 5 '12 at 23:25
    
we can prove that by usisng the same technic (induction principle). –  Abdelmajid Khadari May 5 '12 at 23:29

$$(n+1)^k=\underset{k\text{ factors of }(n+1)}{\underbrace{(n+1)(n+1)\cdots(n+1)}}$$

In expanding that product, the terms will come from choosing either $n$ or $1$ from each of the $k$ factors. If we choose all $1$s, we'll get $n^0$; if we choose all $n$s, we'll get $n^k$; if we choose $p$ $n$s and $k-p$ $1$s (for $0\le p\le k$), we'll get $n^p$. So there are terms with $n^p$ for $p=0,1,\dots,k$, or $k+1$ distinct powers of $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.