Binomial Theorem And Its Simple Applications Question 136
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}= $
[IIT 1966]
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:
- We have $ {{(1+x)}^{15}}=C_0+C_1x+C_2x^{2}.+….+C_{15}x^{15} $
therefore $ \frac{{{(1+x)}^{15}}-1}{x}=C_1+C_2x+….+C_{15}x^{14} $
Differentiating both sides with respect to x, we get $ =\frac{x.15{{(1+x)}^{14}}-{{(1+x)}^{15}}+1}{x^{2}} $ = $ C_2+2C_3x+……+{{,}^{14}}C_{15}x^{13} $
Putting $ x=1 $ , we get $ C_2+2C_3+….+14C_{15}={{15.2}^{14}}-2^{15}+1={{13.2}^{14}}+1. $
BETA
