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On one of my calculus lectures I've seen the lecturer write:

$(1+p)^n=1+np+\frac{n(n-1)}{2}p^2+\cdots+p^n$

Could you please explain to me how did he get this equation?

Thank you very much.

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    Sorry for the question, after applying it with$x=1$and$y=p$I got the same equation. Sorry and thanks again.2012-04-02

2 Answers 2

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You can show the binomial theorem by induction starting from $(1+p)^0=1$ or $(1+p)^1=1+p$

For the abbreviated form you have, you can indicate $(1+p)^n=(1+p)(1+p)^{n-1}$ $ =(1+p)\left(1+(n-1)p+\frac{(n-1)(n-2)}{2}p^2+\cdots+p^{n-1}\right) $ $ =1 +p + (n-1)p+ (n-1)p^2+\frac{(n-1)(n-2)}{2}p^2+\frac{(n-1)(n-2)}{2}p^3+\cdots+p^{n-1}+p^n $ $= 1+np+\frac{n(n-1)}{2}p^2+\cdots+p^n$

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One can also employ generating functions if you already know the power series for $\rm{\it e}^x$

$\begin{eqnarray} \rm {\it e}^{\:(a+b)\:x}\ &=&\rm\ {\it e}^{\:ax}\:{\it e}^{\:bx} \\ \rm \sum \frac{((a+b)\:x)^n}{n!}\ &=&\rm\ \sum \frac{(ax)^n}{n!}\ \sum \frac{(bx)^n}{n!} \\ \rm \frac{(a+b)^n}{n!}\ &=&\rm\ \sum \frac{a^{n-k}}{(n-k)!}\frac{b^k}{k!} \quad by\ comparing\ coef's\ of\ x^n\ above\\ \rm (a+b)^n\ &=&\rm\ \sum \frac{n!}{k!(n-k)!} a^{n-k} b^k\: =\: \sum {n\choose k} a^{n-k} b^k \end{eqnarray}$

The structure at the heart of the above has widespread applications. It is brought to the fore when one studies Umbral Calculus, e.g. see the book by that name by Steven Roman.