Math Doubts

Evaluate $\displaystyle \sum_{n \,=\, 2}^{127} \log_{2} \Bigg(1+\dfrac{1}{n}\Bigg)$

$n$ is a literal number and it represents the numbers from $2$ to $127$. A logarithmic term whose number is sum of $1$ and reciprocal of $n$ and its base $2$. The sum of the binary logarithm terms for the $n$ values start from $2$ to $127$ is symbolically represented by the uppercase sigma symbol.

$\displaystyle \sum_{n \,=\, 2}^{127} \log_{2} \Bigg(1+\dfrac{1}{n}\Bigg)$

The value of the function is required to evaluate in this logarithm problem.

01

Expansion

Substitute $n = 2$ in the logarithmic term and repeat the procedure till $n = 127$ but add all of them to represent the functionality of the uppercase sigma symbol.

$=\,\,\,$ $\log_{2} \Bigg(1+\dfrac{1}{2}\Bigg)$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{3}\Bigg)$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(1+\dfrac{1}{127}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{2+1}{2}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{3+1}{3}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{4+1}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(\dfrac{127+1}{127}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{3}{2}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{4}{3}\Bigg)$ $+$ $\log_{2} \Bigg(\dfrac{5}{4}\Bigg)$ $+$ $\cdots$ $+$ $\log_{2} \Bigg(\dfrac{128}{127}\Bigg)$

02

Merging logarithmic terms

The sum of all logarithmic terms can be merged as a single logarithmic term by using product rule of logarithms.

$=\,\,\,$ $\log_{2} \Bigg[\Bigg(\dfrac{3}{2}\Bigg) \Bigg(\dfrac{4}{3}\Bigg) \Bigg(\dfrac{5}{4}\Bigg) \cdots \Bigg(\dfrac{128}{127}\Bigg)\Bigg]$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{3 \times 4 \times 5 \times \cdots \times 128}{2 \times 3 \times 4 \times \cdots \times 127}\Bigg)$

$=\,\,\,$ $\require{cancel} \log_{2} \Bigg(\dfrac{\cancel{3} \times \cancel{4} \times \cancel{5} \times \cdots \times 128}{2 \times \cancel{3} \times \cancel{4} \times \cdots \times \cancel{127}}\Bigg)$

$=\,\,\,$ $\log_{2} \Bigg(\dfrac{128}{2}\Bigg)$

$=\,\,\,$ $\require{cancel} \log_{2} \Bigg(\dfrac{\cancel{128}}{\cancel{2}}\Bigg)$

$=\,\,\,$ $\log_{2} 64$

03

Finding the value

The base of logarithmic term is $2$. So, express the number in exponential notation on the basis of number $2$.

$=\,\,\,$ $\log_{2} 2^6$

It can be simplified by the power rule of logarithms.

$=\,\,\,$ $6 \times \log_{2} 2$

The logarithm of the base is one as per the logarithm of base rule.

$=\,\,\,$ $6 \times 1$

$=\,\,\,$ $6$

Latest Math Topics
Latest Math Problems
Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved