If we observe closely, we find that the various branches of mathematics are all linked together in some way or the other. I show here one such observation. The binomial theorem can actually be expressed in terms of the derivatives of x^{n} instead of the use of combinations. Lets start with the standard representation of the binomial theorm, We could then rewrite this as a sum, Another way of writing the same thing would be, We observe here that the equation can be rewritten in terms of the derivatives of x^{n}. The coefficient of a in the second terms is the first derivative of x^{n}, similarily the coefficient of a^{2}/2! in the second term is the second derivative of x… Lets look at the last term of the expansion, The coefficient of x^{n}/n! should now be the n^{th} derivative of x^{n}. Which is very true… The following simple relation holds for all n… Hence, the binomial expansion can now be written in terms of derivatives! We have, *where D _{r} represents the rth derivate of x^{n}.* Hence, we can now write this as a sum, http://latex.codecogs.com/gif.latex?huge%20(x+a)^{n}=sum_{0}^{n}frac{D_{r}a^{r}}{r!} So, we now have the expansion in terms of combinations as well as in terms of derivatives!

^{n}instead of the use of combinations. Lets start with the standard representation of the binomial theorm, We could then rewrite this as a sum, Another way of writing the same thing would be, We observe here that the equation can be rewritten in terms of the derivatives of x

^{n}. The coefficient of a in the second terms is the first derivative of x

^{n}, similarily the coefficient of a

^{2}/2! in the second term is the second derivative of x… Lets look at the last term of the expansion, The coefficient of x

^{n}/n! should now be the n

^{th}derivative of x

^{n}. Which is very true… The following simple relation holds for all n… Hence, the binomial expansion can now be written in terms of derivatives! We have,

*where D*Hence, we can now write this as a sum, Or as the sum, So, we now have the expansion in terms of combinations as well as in terms of derivatives!

_{r}represents the rth derivate of x^{n}.
Great. now, can we generate the nth. der, without generating all the derivatives in betweeen

Great work keep it coming