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Find value of A

Albert

Well-known member
Jan 25, 2013
1,225
find value of A:

$A=\dfrac {1+2^{200}+4^{200}+5^{200}+10^{200}+10^{200}+20^{200}+25^{200}+50^{200}+100^{200}} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Hey Albert, I can solve this one! :eek:


$A=\dfrac {1+2^{200}+4^{200}+5^{200}+10^{200}+10^{200}+20^{200}+25^{200}+50^{200}+100^{200}} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$

$A=\dfrac {1+\left(\frac{100}{50}\right)^{200}+\left(\frac{100}{25}\right)^{200}+\left(\frac{100}{20}\right)^{200}+\left(\frac{100}{10}\right)^{200}+\left(\frac{100}{10}\right)^{200}+\left(\frac{100}{5}\right)^{200}+\left(\frac{100}{4}\right)^{200}+\left(\frac{100}{2}\right)^{200}+100^{200}} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$

$A=\dfrac {1+100^{200}\left((\frac{1}{50})^{200}+(\frac{1}{25})^{200}+(\frac{1}{20})^{200}+(\frac{1}{10})^{200}+(\frac{1}{10})^{200}+(\frac{1}{5})^{200}+(\frac{1}{4})^{200}+(\frac{1}{2})^{200}+1\right)} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$

$A=\dfrac {100^{200}\left((\frac{1}{100})^{200}+(\frac{1}{50})^{200}+(\frac{1}{25})^{200}+(\frac{1}{20})^{200}+(\frac{1}{10})^{200}+(\frac{1}{10})^{200}+(\frac{1}{5})^{200}+(\frac{1}{4})^{200}+(\frac{1}{2})^{200}+1\right)} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$

$A=\dfrac {100^{200}\left(1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}\right)} {1+2^{-200}+4^{-200}+5^{-200}+10^{-200}+10^{-200}+20^{-200}+25^{-200}+50^{-200}+100^{-200}}$

$A=100^{200}$
 

Albert

Well-known member
Jan 25, 2013
1,225
anemone :
perfect ! you got the answer (Yes)