# $(n + 1)^k$ number of different degrees of $n$

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

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

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 $k+1$.

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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 $n$.

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