Question

$\dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{{{4.2}^4}}} + ......$ is equal to
A) $\dfrac{1}{4}$
B) ${\log _e}\dfrac{3}{4}$
C) ${\log _e}\dfrac{3}{2}$
D) ${\log _e}\dfrac{2}{3}$

Answer 1

Hint: We have to find the sum of $\dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{{{4.2}^4}}} + ......$. We will find the sum of given series using expansion of $\log $function. We will use the formula of $\log (1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ......$

Complete step by step Solution:
Given, $\dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{{{4.2}^4}}} + ......$
We have to find the sum of given infinite series that is $\dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{{{4.2}^4}}} + ......$
Let $A = \dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{{{4.2}^4}}} + ......$
To sum the series we will use the logarithm function.
We know that $\log (1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ......$
Let $x = \dfrac{1}{2}$
$\log (1 + \dfrac{1}{2}) = \dfrac{1}{2} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^2}}}{2} + \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^3}}}{3} - \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^4}}}{4} + ......$
${\log _e}(1 + \dfrac{1}{2}) = \dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{4.2{}^4}} + ......$
${\log _e}\left( {\dfrac{3}{2}} \right) = \dfrac{1}{2} - \dfrac{1}{{{{2.2}^2}}} + \dfrac{1}{{{{3.2}^3}}} - \dfrac{1}{{4.2{}^4}} + ......$

Hence, the correct option is (c).

Additional Information:
In arithmetic, before the invention of calculus, several maths students used logarithms to vary multiplication and division issues into addition and subtraction issues. In Logarithms, the ability is raised to some numbers (usually, base variety) to urge another number. it associates mathematical functions with mathematical functions. We all know that arithmetic and Science perpetually upset the big powers of numbers, logarithms square measure most significant and helpful.
Logarithmic functions in arithmetic are associate operators which are able to assist you to specifically calculate the exponent which will satisfy the equation. The index operation is that the inverse of the exponential function A function in arithmetic in a very relationship between a gaggle of inputs specified every has one output. In easy terms, it's a relationship between inputs during which every input is related to only 1 output.

Note: Students should be more careful with the formula if they choose the wrong formula they will not get the correct answer. Students can also make mistakes while applying formulas. So, they should pay attention while putting the value of x to avoid calculation errors.