Binomial Theorem And Its Simple Applications Question 173
Question: $ 2C_0+\frac{2^{2}}{2}C_1+\frac{2^{3}}{3}C_2+….+\frac{2^{11}}{11}C_{10} $
[MP PET 1999; EAMCET 1992]
Options:
A) $ \frac{3^{11}-1}{11} $
B) $ \frac{2^{11}-1}{11} $
C) $ \frac{11^{3}-1}{11} $
D) $ \frac{11^{2}-1}{11} $
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ {{(1+x)}^{10}}=C_0+C_1x+C_2x^{2}+…+C_{10}x^{10} $ Integrating both sides from 0 to 2, we get $ \frac{3^{11}-1}{11}=2C_0+\frac{2^{2}}{2}C_1+\frac{2^{3}}{3}C_2+….+\frac{2^{11}}{11}C_{10} $ .
BETA
