Question

how to solve this problem​

Answer 1

Answer:

$\frac{2019}{1010}$

Step-by-step explanation:

Given :

To find the value of :$\frac{\frac{1}{2}}{1 + \frac{1}{2}}+\frac{\frac{1}{3} }{(1+\frac{1}{2})(1+\frac{1}{3})} + ... + \frac{\frac{1}{2019}}{(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{2019})}$

Solution :

The given summation,

$\frac{\frac{1}{2}}{1 + \frac{1}{2}}+\frac{\frac{1}{3} }{(1+\frac{1}{2})(1+\frac{1}{3})} + ... + \frac{\frac{1}{2019}}{(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{2019})}$

Remember,

i) $1+\frac{1}{2} = \frac{3}{2}$  = 1.5

ii) $1+\frac{1}{3} = \frac{4}{3}$

⇒ $(1+\frac{1}{2})(1+\frac{1}{3}) = (\frac{3}{2})(\frac{4}{3})$ = 2

iii) $1+\frac{1}{4} = \frac{5}{4}$

⇒ $(1+\frac{1}{2})(1+\frac{1}{3})(1 +\frac{1}{4}) = (\frac{3}{2})(\frac{4}{3})(\frac{5}{4}) = \frac{5}{2}$ = 2.5

Hence,

The series goes on as , 1.5 , 2 , 2.5 , 3 ,...

(or) $\frac{3}{2} , \frac{4}{2} , \frac{5}{2} , \frac{6}{2} ,...$

The summation can also be written as,

$\frac{\frac{1}{2}}{\frac{3}{2}} + \frac{\frac{1}{3}}{\frac{4}{2}} + \frac{\frac{1}{4}}{\frac{5}{2}} +...+\frac{\frac{1}{2019}}{\frac{2020}{2}}$

⇒ $(\frac{1}{2} \times \frac{2}{3}) +(\frac{1}{3} \times \frac{4}{2}) + (\frac{1}{4} \times \frac{2}{5}) +...+(\frac{1}{2019} \times \frac{2}{2020} )$

⇒ $\frac{2}{2\times3}+\frac{2}{3\times4} + \frac{2}{4\times5} + .... + \frac{2}{2019\times2020}$

⇒ $2(\frac{1}{2(2+1)} + \frac{1}{3(3+1)} + \frac{1}{4(4+1)} +...+\frac{1}{2019(2019+1)})$ ...(i)

Note :

We notice that,

$\frac{1}{n} - \frac{1}{n+1} = \frac{1(n+1) - 1(n)}{n(n+1)} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)}$

By using this algoritm in (i),

We get,.

⇒ $2\left[\left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) +...+ \left(\frac{1}{2018} - \frac{1}{2019}\right) + \left(\frac{1}{2019} - \frac{1}{2020}\right)\right]$

⇒ $2(1 - \frac{1}{2020}) = 2(\frac{2020 - 1}{2020}) = 2(\frac{2019}{2020})=\frac{4038}{2020}=\frac{2019}{2020}=1\frac{1009}{1010}$

∴ The value of $\frac{\frac{1}{2}}{1 + \frac{1}{2}}+\frac{\frac{1}{3} }{(1+\frac{1}{2})(1+\frac{1}{3})} + ... + \frac{\frac{1}{2019}}{(1+\frac{1}{2})(1+\frac{1}{3})...(1+\frac{1}{2019})} =\frac{2019}{1010}$