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Calculus Level 3

$\dfrac{a-b}{a}+\dfrac{1}{2} \left(\dfrac{a-b}{a} \right)^2+\dfrac{1}{3} \left(\dfrac{a-b}{a} \right)^3+ \ldots$

If $0 < b < a$ , the value of the above series is?

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None of the given choices

\ln \left( \frac{b}{a}\right)

\ln \left( \frac{a}{b}\right)

1

1 solution

Tanishq Varshney
Mar 29, 2015

$-ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+......$

here $x=\frac{a-b}{a}$

= $-ln(\frac{b}{a})=ln(\frac{a}{b})$

For to proove it ....

$\displaystyle{f\left( x \right) =\sum _{ k=1 }^{ \infty }{ \cfrac { { x }^{ k } }{ k } } =x+\cfrac { { x }^{ 2 } }{ 2 } +........\\ diff.\quad w.r.t\quad x\\ f^{ ' }\left( x \right) =\sum _{ k=1 }^{ \infty }{ { x }^{ k-1 } } =1+x+{ x }^{ 2 }+{ x }^{ 3 }+.......\quad (GP)\\ f^{ ' }\left( x \right) =\cfrac { 1 }{ 1-x } \\ int.\quad w.r.t\quad x\\ f(x)=\int { f^{ ' }\left( x \right) dx } =\int { \cfrac { 1 }{ 1-x } dx } =-\ln { (1-x) } +c\\ f(0)=0\\ f(x)=-\ln { (1-x) } \\ \boxed { f(\cfrac { a-b }{ a } )=\ln { \cfrac { a }{ b } } } }$

Deepanshu Gupta - 6 years, 2 months ago

Did it the same way ! Learnt so much on brilliant. Thanks !

Keshav Tiwari - 6 years, 2 months ago

i learnt in fiitjee and also on brilliant but fiitjee more

Shashank Rustagi - 6 years, 2 months ago

@Sandeep Bhardwaj I have added the condition that $0 < b < a$ .

Calvin Lin Staff - 6 years, 2 months ago

Thanks sir.

Sandeep Bhardwaj - 6 years, 2 months ago

Déjà vu

Try the set: Practice Problems for JEE-Mains and Give a final Boost to your JEE_preparation. All the best!

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