Binomial Theorem And Its Simple Applications Question 117
Question: The coefficient of $ x^{83} $ in $ {{(1+x+x^{2}+x^{3}+x^{4})}^{n}} $ $ {{(1-x)}^{n+3}},is-{{,}^{n}}{C_{2\lambda }} $ , then find the value of $ \lambda $
Options:
A) 12
B) 10
C) 9
D) 8
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] We have, $ {{( 1+x+x^{2}+x^{3}+x^{4} )}^{n}}{{(x-1)}^{n+3}} $
$ =\frac{{{(x^{5}-1)}^{n}}}{{{(1-x)}^{n}}}.{{(x-1)}^{n+3}}={{(x^{5}-1)}^{n}}{{(x-1)}^{3}} $
$ =( +x^{3}-3x^{2}+3x-1 )\sum\limits_{r=0}^{n}{^{n}C_{r}{{(-1)}^{r}},x^{5r}} $
$ =+\sum\limits_{r=0}^{n}{{{,}^{n}}C_{r}{{(-1)}^{r}}{x^{5r+3}}+3\sum\limits_{r=0}^{n}{{{,}^{n}}C_{r}{{(-1)}^{r}},{x^{5r+2}}}} $
$ -3\sum\limits_{r=0}^{n}{{{,}^{n}}C_{r}{{(-1)}^{r}},{x^{5r+1}}+3\sum\limits_{r=0}^{n}{{{,}^{n}}C_{r}}{{(-1)}^{r}}x^{5r}} $
For term containing $ x^{83}, $ we have $ 5r+3=83 $
$ \Rightarrow r=16 $ whereas $ 5r+2=83,5r+1=83 $ and $ 5r=83 $ give no integral value of r.
Hence, there is only one term containing $ x^{83} $ whose coefficient $ =-{{,}^{n}}C_{16}=-{{,}^{n}}{C_{2\lambda }},\therefore 2\lambda =16\Rightarrow \lambda =8 $
BETA
