qui peux m'aider à faire ce dm merci.... Pergunta de ideia deanniebusschaert

June 2021 1 128 Report

qui peux m'aider à faire ce dm
merci

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Amateur Bonjour,

Joli comme raisonnement. Compliments au professeur de PRIMAIRE.
$1)\\
A= \dfrac{1}{1} +\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\\

= (\dfrac{1}{1} +\dfrac{1}{2})+\dfrac{1}{3}+\dfrac{1}{4}\\

= (\dfrac{2}{1} +\dfrac{1}{2})+\dfrac{1}{3}+\dfrac{1}{4}\\

= \dfrac{3}{2} +\dfrac{1}{3}+\dfrac{1}{4}\\

= (\dfrac{3}{2} +\dfrac{1}{3})+\dfrac{1}{4}\\

= (\dfrac{9}{6} +\dfrac{2}{6})+\dfrac{1}{4}\\

= \dfrac{11}{6} +\dfrac{1}{4}\\

= \dfrac{22}{12} +\dfrac{3}{12}\\

= \dfrac{25}{12}\\

2)\\
B= A+\dfrac{1}{5} = \dfrac{25}{12} +\dfrac{1}{5}= \dfrac{125}{60} +\dfrac{12}
{60}\\



$

=137/60

3-4)
$\sum_{i=1}^5( \dfrac{1}{i} )= \frac{137}{60} \\


 \dfrac{1}{ \sum_{i=1}^5( i)}= \dfrac{1}{15} \\


$

II)
1-2)
$1^2+2^2+3^2=\sum_{i=1}^3 (i^2)=1+4+9=14\\

\sum_{i=1}^{100} (i^2)= \dfrac{100*(100+1)*(2*100+1)}{6} \\
=\dfrac{(100*101*201)}{6}=338350\\
$

3)
$\sum_{i=1}^{n} (i^2)= \dfrac{n*(n+1)*(2*n+1)}{6}\\

= \dfrac{(n^2+n)*(2*n+1)}{6}\\

= \dfrac{n^3+3n^2+n}{6}\\

= \dfrac{n^3}{6}+ \dfrac{n^2}{2}+ \dfrac{n}{6}\\
$

III)
1)
$1*2*3*4+1=24+1=25=5^2\\
2*3*4*5+1=120+1=121=11^2\\
3*4*5*6+1=360+1=361=19^2\\

$

2) Bien vu Tryphon.

3)
a)
$(n+1)(n+2)=n^2+n+2n+2=n^2+3n+2=n(n+3)+2\\

b)\\

a=(n+1)(n+2)=n(n+3)+2\ \Longrightarrow a-2=n(n+3)\\

p+1=n(n+1)(n+2)(n+3)+1=n(n+3)(a-2)=(a-2)*(a-2)=(a-2)^2\\




$

4) La démonstration est en 3)

1 votes Thanks 0

anniebusschaert merci beaucoup

Amateur de rien

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