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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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As your title indicates, he used the Binomial Theorem. The link supplies a proof. It is the proof that you have difficulty with? –  André Nicolas Apr 2 '12 at 21:46
    
yes, I guesses he used the Binomial Theorem, but doesn't the Binomial Theorem looks a bit different than this? –  Anonymous Apr 2 '12 at 21:48
    
If you expand the expressions on the Binomial theorem, you should get those numbers. –  Marra Apr 2 '12 at 21:49
    
What does the binomial theorem look like if you spell out each coefficient instead of using fancy notation like $\binom{n}{i}$? –  Dilip Sarwate Apr 2 '12 at 21:49
    
@Anonymous: Different? –  Salech Alhasov Apr 2 '12 at 21:49
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2 Answers

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.

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