Simple Question on Binomial theorems...

Question

I have tried to solve this question by putting the value of each coefficients but it is really becoming very lengthy.... what i got was this number 10510100501... But how to get this in the required form?? Please help me... Also is there any other elegant way to solve this question...Any help would be appreciated.. Moreover what i had figured out was that the each numeric terms that are premultiplied are actually the coefficients of the binomial expansion of (1-x)^5.... Thanks in advance...........$(value of n is 101)$..

Answer 1

$$\begin{align}&\binom{505}{5}-5\binom{404}{5}+10\binom{303}{5}-10\binom{202}{5}+5\binom{101}{5}\\&=\sum_{k=1}^{5}(-1)^{k-1}\binom{5}{k}\binom{101k}{5}\\&=\sum_{k=0}^{5}(-1)^{k-1}\binom{5}{k}\binom{101k}{5}\\&=-\sum_{k=0}^{5}(-1)^{k}\binom{5}{k}\binom{101k}{5}\\&=-\sum_{k=0}^{5}(-1)^{k}\binom 5k(1+x)^{101k}\qquad [x^5]\\&=-\sum_{k=0}^{5}\binom 5k\cdot 1^{5-k}\cdot(-(1+x)^{101})^k\qquad [x^5]\\&=-(1-(1+x)^{101})^5\qquad [x^5]\\&=-(1-(1+x))^5(1+(1-x)+(1-x)^2+\cdots +(1-x)^{100})^5\qquad [x^5]\\&=(1+(1-x)+(1-x)^2+\cdots +(1-x)^{100})^5\qquad [x^0]\\&=101^5\end{align}$$