Binomial Theorem And Its Simple Applications Question 43
Question: $ \frac{1}{2}x^{2}+\frac{2}{3}x^{3}+\frac{3}{4}x^{4}+\frac{4}{5}x^{5}+ $ …………….. is
Options:
A) $ \frac{x}{1+x}+\log (1+x) $
B) $ \frac{x}{1-x}+\log (1+x) $
C) $ -\frac{x}{1-x}+\log (1+x) $
D) $ \frac{x}{1-x}+\log (1-x) $
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Answer:
Correct Answer: D
Solution:
- [d] $ \frac{1}{2}x^{2}+\frac{2}{3}x^{3}+\frac{3}{4}x^{4}+\frac{4}{5}x^{5}+….. $
$ =( 1-\frac{1}{2} )x^{2}+( 1-\frac{1}{3} )x^{3}+( 1-\frac{1}{4} )x^{4}+ $
$ ( 1-\frac{1}{5} )x^{5}+… $
$ =(x^{2}+x^{3}+x^{4}+x^{5}+……)+ $
$ ( -\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}…… ) $
$ =(x+x^{2}+x^{3}+…)+( -\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}…… ) $
$ =\frac{x}{1-x}+{\log_{e}}(1-x) $
BETA
