Binomial Theorem And Its Simple Applications Question 32
Question: If $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+……….+C_{n}x^{2}, $ then $ C_0^{2}+C_1^{2}+C_2^{2}+C_3^{2}+……+C_n^{2} $ =
[MP PET 1985; Karnataka CET 1995; MNR 1999]
Options:
A) $ \frac{n!}{n!n!} $
B) $ \frac{(2n)!}{n!n!} $
C) $ \frac{(2n)!}{n!} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {{(1+x)}^{n}}=C_0+C_1x+C_2x^{2}+…..+C_{n}x^{n} $ …..(i) and $ {{( 1+\frac{1}{x} )}^{n}}=C_0+C_1\frac{1}{x}+C_2{{( \frac{1}{x} )}^{2}}+…..+C_{n}{{( \frac{1}{x} )}^{n}} $ ….(ii)
If we multiply (i) and (ii), we get $ C_0^{2}+C_1^{2}+C_2^{2}+…..+C_n^{2} $ is the term independent of x and
Hence it is equal to the term independent of x in the product $ {{(1+x)}^{n}}{{( 1+\frac{1}{x} )}^{n}} $ or in $ \frac{1}{x^{n}}{{(1+x)}^{2n}} $ or term containing $ x^{n} $ in $ {{(1+x)}^{2n}} $ .
Clearly the coefficient of $ x^{n} $ in $ {{(1+x)}^{2n}} $ is $ {T_{n+1}} $ and equal to $ ^{2n}C_{n}=\frac{(2n)!}{n!n!} $ Trick :
Solving conversely. Put $ n=1,n=2,….. $ then we get $ S_1={{,}^{1}}C_0^{2}+{{,}^{1}}C_1^{2}=2 $ , $ S_2={{,}^{2}}C_0^{2}+{{,}^{2}}C_1^{2}+{{,}^{2}}C_2^{2}=1^{2}+2^{2}+1^{2}=6 $
Now check the options (a) Does not hold given condition, (b) (i) Put $ n=1 $ , then $ \frac{2!}{1!,1!}=2 $ (ii) Put $ n=2 $ , then $ \frac{4!}{2!,2!}=\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=6 $
Note: Students should remember this question as an identity.
BETA
