Question

The value of $x$ satisfying the equation $\frac{5050-(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+......+\frac{5049}{5050})}{1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{5050}}=\frac{x}{5050}$ is

• A. $1$
• B. $5049$
• C. $5050$
• D. $5051$

Answer 1

Answer: (C)

$5050$

First we solve,

$(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+.......+\frac{5049}{5050})$

We can conclude general term or $n^{th}$ term,

$={\sum }_{n=1}^{5049}\frac{n}{n+1}$

$={\sum }_{n=1}^{5049}\frac{n+1-1}{n+1}$

$={\sum }_{n=1}^{5049}(1-\frac{1}{n+1})$

$=(1-\frac{1}{1+1})+(1-\frac{1}{2+1})+.......+(1-\frac{1}{5049+1})$

$=(1-\frac{1}{2})+(1-\frac{1}{3})+.......+(1-\frac{1}{5050})$

$=5049-(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{5050})$

Substituting this value in $LHS$ of the given expression,

$\frac{5050-(\frac{1}{2}+\frac{2}{3}+.......+\frac{5049}{5050})}{1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{5050}}$

$=\frac{5050-5049+(\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{5050})}{1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{5050}}$

$=\frac{1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{5050}}{1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{5050}}$

$=1$

Now,

$\frac{x}{5050}=1$

$x=5050$

This is the required value of $x$.