If
$$A = \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+...+\frac{1}{1+50}$$
dan
$$B = \frac{1}{1+1}+\frac{1}{1+\frac{1}{2}}+\frac{1}{1+\frac{1}{3}}+...+\frac{1}{1+\frac{1}{50}}$$
then the value of $5A^2+5B^2+10AB=...$

Question

If 
$$A = \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+...+\frac{1}{1+50}$$ 
dan 
$$B = \frac{1}{1+1}+\frac{1}{1+\frac{1}{2}}+\frac{1}{1+\frac{1}{3}}+...+\frac{1}{1+\frac{1}{50}}$$ 
then the value of $5A^2+5B^2+10AB=...$

mathslover · Answer

So, basically, we have to find : $5(A+B)^2$ 

Right?

chihiroasleaf · Answer

yes.., but I have no  idea how to simplify A and B

mathslover · Answer

Now, try this! A+ B

mathslover · Answer

Well, one minute. Writing LaTeX

amistre64 · Answer

is the latex coding for yall?

mathslover · Answer

Yes!

amistre64 · Answer

must be my computer then

vishweshshrimali5 · Answer

Well, let @mathslover solve it... then I will show you a trick ;)

mathslover · Answer

I will represent terms of A in red and that of B in blue 

$\color{red}{A} + \color{blue}{B} = \left( \color{red}{\cfrac{1}{1+1}} + \color{blue}{\cfrac{1}{1+1}} \right) + \left( \color{Red}{\cfrac{1}{1+2}} + \color{blue}{\cfrac{1}{1+1/2}} \right) + ... + \left( \color{red}{\cfrac{1}{1+50}} + \color{blue}{\cfrac{1}{1+1/50}} \right) $

vishweshshrimali5 · Answer

Well, looks like that he caught it :D

chihiroasleaf · Answer

ahhh..., I get it... :))

chihiroasleaf · Answer

A+B = 51 ?

mathslover · Answer

$\color{red}{A} + \color{blue}{B} = \left( \color{red}{\cfrac{1}{1+1}} + \color{blue}{\cfrac{1}{1+1}} \right) + \left( \color{Red}{\cfrac{1}{1+2}} + \color{blue}{\cfrac{1}{1+1/2}} \right) + ... \ + \left( \color{red}{\cfrac{1}{1+50}} + \color{blue}{\cfrac{1}{1+1/50}} \right) $

mathslover · Answer

Now, we have : 

$A + B = 1 + 1 + ... + 1$ 

Clearly it is 51 :)

mathslover · Answer

This is how I came up with that : 

$\cfrac{1}{1+1} + \cfrac{1}{1+1} = 1$
$\cfrac{1}{1+2} + \cfrac{1}{1 +1/2} = \cfrac{1}{3} + \cfrac{2}{3} = 1$ 

And so on..

So you get 1 + 1 + .. + 1 (50 times) = 50 

(Sorry for the mistake earlier, it is 50 not 51)

mathslover · Answer

@vishweshshrimali5  - What was the trick you were talking about? Am curious to know about it

chihiroasleaf · Answer

ahhh..., I got the same mistake.., 
yes.., it's 50 

so, $5(50)^2 $

vishweshshrimali5 · Answer

Well... I really like using $\Sigma$

So:

$$\large{A = \Sigma_{i = 1}^{50} \cfrac{1}{1+i}}$$
$$\large{B = \Sigma_{i = 1}^{50} \cfrac{1}{1+\cfrac{1}{i}}} = \Sigma \cfrac{i}{1+i}$$

$$\large{A+B = \Sigma_{i=1}^{50} 1 = 50}$$

Thus: $\large{5(A+B)^2 = 5 *(50)^2}$