Binomial Theorem And Its Simple Applications Question 50
Question: The coefficients of three successive terms in the expansion of $ {{(1+x)}^{n}} $ are 165, 330 and 462 respectively, then the value of n will be
[UPSEAT 1999]
Options:
A) 11
B) 10
C) 12
D) 8
Show Answer
Answer:
Correct Answer: A
Solution:
- Let the coefficient of three consecutive terms i.e. $ {{(r+1)}^{th}},{{(r+2)}^{th}},,{{(r+3)}^{th}} $ in expansion of $ {{(1+x)}^{n}} $ are 165,330 and 462 respectively then, coefficient of $ {{(r+1)}^{th}} $ term $ ={}^{n}C_{r}=165 $ Coefficient of (r + 2)th term $ ={}^{n}{C_{r+1}}=330 $ and Coefficient of (r + 3)th term $ ={}^{n}{C_{r+2}}=462 $ \ $ \frac{{}^{n}{C_{r+1}}}{{}^{n}C_{r}}=\frac{n-r}{r+1}=2 $ or $ n-r=2(r+1) $ or $ r=\frac{1}{3}(n-2) $ and $ \frac{{}^{n}{C_{r+2}}}{{}^{n}{C_{r+1}}}=\frac{n-r-1}{r+2}=\frac{231}{165} $ or $ 165(n-r-1)=231(r+2) $ or $ 165n-627=396r $ or $ 165n-627=396\times \frac{1}{3}\times (n-2) $ or $ 165n-627=132(n-2) $ or n = 11.
BETA
