Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Calculate the sum of the series$\dfrac{1}{{{{\log }_2}4}} + \dfrac{1}{{{{\log }_4}4}} + \dfrac{1}{{{{\log }_8}4}} + ............................ + \dfrac{1}{{{{\log }_{{2^n}}}4}}$$\left( A \right)\dfrac{{n\left( {n + 1} \right)}}{2}$$\left( B \right)\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{2}$$\left( C \right)\dfrac{1}{{n\left( {n + 1} \right)}}$$\left( D \right)\dfrac{{n\left( {n + 1} \right)}}{4}$

Last updated date: 09th Aug 2024
Total views: 69k
Views today: 0.69k
Verified
69k+ views
Hint – In this particular question use the property of logarithmic (i.e. ${\log _a}b = \dfrac{{\log b}}{{\log a}}$, $\log {a^b} = b\log a$) and use the direct formula of the sum of first natural numbers (i.e. $\dfrac{{n\left( {n + 1} \right)}}{2}$) so use these concepts to reach the solution of the question.

Given series
$\dfrac{1}{{{{\log }_2}4}} + \dfrac{1}{{{{\log }_4}4}} + \dfrac{1}{{{{\log }_8}4}} + ............................ + \dfrac{1}{{{{\log }_{{2^n}}}4}}$
Now this equation is also written as
$\Rightarrow \dfrac{1}{{{{\log }_2}4}} + \dfrac{1}{{{{\log }_{{2^2}}}4}} + \dfrac{1}{{{{\log }_{{2^3}}}4}} + ............................ + \dfrac{1}{{{{\log }_{{2^n}}}4}}$
Now use logarithmic property i.e. ${\log _a}b = \dfrac{{\log b}}{{\log a}}$, so use this property in the above equation we have,
$\Rightarrow \dfrac{{\log 2}}{{\log 4}} + \dfrac{{\log {2^2}}}{{\log 4}} + \dfrac{{\log {2^3}}}{{\log 4}} + ............................ + \dfrac{{\log {2^n}}}{{\log 4}}$
Now as we know that 4 is written as ${2^2}$ so we have,
$\Rightarrow \dfrac{{\log 2}}{{\log {2^2}}} + \dfrac{{\log {2^2}}}{{\log {2^2}}} + \dfrac{{\log {2^3}}}{{\log {2^2}}} + ............................ + \dfrac{{\log {2^n}}}{{\log {2^2}}}$
Now again use logarithmic property i.e. $\log {a^b} = b\log a$, so use this property in the above equation we have,
$\Rightarrow \dfrac{{\log 2}}{{2\log 2}} + \dfrac{{2\log 2}}{{2\log 2}} + \dfrac{{3\log 2}}{{2\log 2}} + ............................ + \dfrac{{n\log 2}}{{2\log 2}}$
Now cancel out log2 from numerator and denominator we have,
$\Rightarrow \dfrac{1}{2} + \dfrac{2}{2} + \dfrac{3}{2} + ............................ + \dfrac{n}{2}$
$\Rightarrow \dfrac{1}{2}\left[ {1 + 2 + 3 + ............................ + n} \right]$
Now as we know that the sum of first natural numbers is given as, i.e. (1 + 2 + 3 + ........... + n) = $\dfrac{{n\left( {n + 1} \right)}}{2}$ so use this property in the above equation we have,
$\Rightarrow \dfrac{1}{2}\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]$
Now simplify this we have,
$\Rightarrow \dfrac{{n\left( {n + 1} \right)}}{4}$
So this is the required sum of the given series.
So this is the required answer.
Hence option (D) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall all the properties of logarithmic and the sum of (n) natural numbers which is all stated above these formula are really helpful to solve these kinds of problems, so first use the logarithmic property and simplify the given series then apply the formula of the sum of first natural numbers and again simply as above we will get the required answer which is option (D).