Binomial Theorem And Its Simple Applications Question 112
Question: If $ {{(1+x)}^{15}}=C_0+C_1x+C_2x^{2}+…+C_{15}x^{15} $ then $ C_2+2C_3+3C_4+…..+14C_{15} $ is equal to
Options:
A) $ {{14.2}^{14}} $
B) $ {{13.2}^{14}}+1 $
C) $ {{13.2}^{14}}-1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] The general term $ T_{r}=(r-1){{,}^{15}}C_{r}, $
$ r=2,3,4,…..,15 $
$ \therefore T_{r}=r{{,}^{15}}C_{r}-{{,}^{15}}C_{r} $
$ =15.{{,}^{14}}{C_{r-1}}-{{,}^{15}}C_{r},[\because ,r.{{,}^{n}}C_{r}=n{{.}^{n-1}}{C_{r-1}}] $
$ \therefore \sum\limits_{r=2}^{15}{T_{r}=15[ ^{14}C_1{{+}^{14}}C_2+….+{{,}^{14}}C_{14} ]} $
$ -[ ^{15}C_2+{{,}^{15}}C_3+…+{{,}^{15}}C_{15} ] $
$ =15[2^{14}-1]-[2^{15}-1-15] $
$ =(15-2)2^{14}+1={{13.2}^{14}}+1 $
BETA
