Binomial Theorem And Its Simple Applications Question 198
Question: $ C_0C_{r}+C_1{C_{r+1}}+C_2{C_{r+2}}+….+{C_{n-r}}C_{n} $ =
[BIT Ranchi 1986]
Options:
A) $ \frac{(2n)!}{(n-r),!,(n+r)!} $
B) $ \frac{n!}{(-r)!(n+r)!} $
C) $ \frac{n!}{(n-r)!} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
*$ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+….+C_{r}x^{r}+…. $ …………..(i)
$ {{( 1+\frac{1}{x} )}^{n}}=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^{2}}+…..+C_{r}\frac{1}{x^{r}}+…. $ …………..(ii)
Multiplying both sides and equating coefficient of $ x^{r} $ in $ \frac{1}{x^{n}}{{(1+x)}^{2n}} $ or the coefficient of $ {x^{n+r}} $ in $ {{(1+x)}^{2n}} $ we get the value of required expression = $ ^{2n}{C_{n+r}}=\frac{(2n),!}{(n-r),!,(n+r),!} $
Trick:
Solving conversely. Put $ n=1 $ and $ r=0 $ in first term , (given condition) (i) $ ^{1}C_0^{1}C_0{{+}^{1}}C_1^{1}C_1=1+1=2 $ , $ (\because r\le n) $ Put $ n=2,r=1 $ , then (ii) $ ^{2}C_0^{2}C_1{{+}^{2}}C_1^{2}C_2=2+2=4 $
Now check the options (i) Put $ n=1,r=0 $ , we get $ \frac{2!}{(1)!(1)!}=2 $ (ii) Put $ n=2,r=1 $ , we get
$ \frac{4!}{(1),!(3),!}=4 $
BETA
