w2m=1(2m)+1+1(2m)+2+..+12m+1w_{2^m} = \frac {1}{(2^m) + 1 } + \frac {1}{(2^m) + 2 } +..+ \frac {1}{2^{m+1}}w2m=(2m)+11+(2m)+21+..+2m+11
Et U(U(U({2m+1})−U</em>2m)-U</em>{2m})−U</em>2m = 12m+1+12m+2+..+12m+1\frac {1}{2^m +1 } + \frac {1}{2^m +2} + .. + \frac {1}{2^{m+1} }2m+11+2m+21+..+2m+11
Voila , donc W2mW_{2m}W2m = UU(+1)U_{U(+1)}UU(+1) - Un .
Donc , apres avoir fais cela , en deduire :
b. Deduire que :
w2m+1=1+∑k=1mw2kw_{2^{m+1} }= 1 + \displaystyle \sum_{k=1}^{m} w_{2^k}w2m+1=1+k=1∑mw2k
Donc :
w2k=∑p=2k+12k+1=12k+1+12k+2+..+12k+1w_{2^k}= \displaystyle \sum_{p=2^k +1}^{2^{k+1}} = \frac {1}{2^k +1} + \frac {1}{2^k +2} + .. + \frac {1}{2^{k+1}}w2k=p=2k+1∑2k+1=2k+11+2k+21+..+2k+11
Donc :
∑k=1mw2k=(121+1+121+2+..+121+1)+(122+1+122+2+..+122+1)+(123+1+123+2+..+123+1)+..+(12m+1+12m+2+..+12m+1)\displaystyle \sum_{k=1}^{m} w_{2^k}= ( \frac {1}{2^1+1} + \frac {1}{2^1+2} + .. + \frac {1}{2^{1+1}}) + ( \frac {1}{2^2+1} + \frac {1}{2^2+2} + .. + \frac {1}{2^{2+1}}) + ( \frac {1}{2^3+1} + \frac {1}{2^3+2} + .. + \frac {1}{2^{3+1}}) + .. + (\frac {1}{2^m+1} + \frac {1}{2^m+2} + .. + \frac {1}{2^{m+1}})k=1∑mw2k=(21+11+21+21+..+21+11)+(22+11+22+21+..+22+11)+(23+11+23+21+..+23+11)+..+(2m+11+2m+21+..+2m+11)
=(13+14)+(15+16+17+18)+(19+110)+(111+112+...+116)+(12m+1+12m+2+..+12m+1)=13+14+15+...+12m+1)= ( \frac {1}{3} + \frac {1}{4}) + ( \frac {1}{5} + \frac {1}{6} + \frac {1}{7} + \frac {1}{8})+ (\frac {1}{9} + \frac {1}{10}) + ( \frac {1}{11} + \frac {1}{12} + ... + \frac {1}{16} ) + (\frac {1}{2^m+1} + \frac {1}{2^m+2} + .. + \frac {1}{2^{m+1}}) = \frac {1}{3} + \frac {1}{4} + \frac {1}{5} +... + \frac {1}{2^{m+1}})=(31+41)+(51+61+71+81)+(91+101)+(111+121+...+161)+(2m+11+2m+21+..+2m+11)=31+41+51+...+2m+11)
Donc :
1+∑k=1mw2k=1+13+14+15+...+12m+11 + \displaystyle \sum_{k=1}^{m} w_{2^k} = 1 + \frac {1}{3} + \frac {1}{4} + \frac {1}{5} +... + \frac {1}{2^{m+1}}1+k=1∑mw2k=1+31+41+51+...+2m+11
Et u2m+1=1+12+13+14+15+...+12m+1u_{2^{m+1}} = 1 + \frac {1}{2} + \frac {1}{3} + \frac {1}{4} + \frac {1}{5} +... + \frac {1}{2^{m+1}}u2m+1=1+21+31+41+51+...+2m+11
Donc , je remarque qu'il manque un '1/2' , dans 1+∑k=1mw2k1 + \displaystyle \sum_{k=1}^{m} w_{2^k}1+k=1∑mw2k , ou est mon erreur ? ! je ne la voi pas .
Merci beaucoup !
ps : le plus dure dans l'exercice , c'est de l'ecrire avec Latex