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julien778899
@julien778899
June 2021
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Suite géométrique et arithmétique SVP aidez moi
merci
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caylus
Verified answer
Bonsoir,
1) voir fichier joint
2)
Comme V[n]=n(n+1)/2
V[n-1]=(n-1)*n/2
V[n-1]+n=(n²-n)/2+2n/2=(n²+n)/2=n(n+1)/2=V[n]
3)W[n]=(V[n])²={n(n+1)/2}²=n²(n²+2n+1)/4=(n^4+2n^3+n²)/4
4*(V[n])²+(n+1)^3)=n^4+2n^3+n²+4n^3+12n²+12n+4
=n^4+6n^3+13n²+12n+4
=n^4+2n^3+4n^3+8n²+5n²+10n+2n+4
=n^3(n+2)+4n²(n+2)+5n(n+2)+2(n+2)
=(n+2)(n^3+4n²+5n+2)=(n+2)(n^3+2n^2+2n^2+4n+n+2)
=(n+2)[n²(n+2)+2n(n+2)+1(n+2)]
=(n+2)(n+2)(n²+2n+1)
=(n+1)²*(n+2)²
=> (V[n])²+(n+1)²={(n+1)/n+2)/2}²={V[n+1]}²
W[n+1]={V[n+1]}²
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Verified answer
Bonsoir,1) voir fichier joint
2)
Comme V[n]=n(n+1)/2
V[n-1]=(n-1)*n/2
V[n-1]+n=(n²-n)/2+2n/2=(n²+n)/2=n(n+1)/2=V[n]
3)W[n]=(V[n])²={n(n+1)/2}²=n²(n²+2n+1)/4=(n^4+2n^3+n²)/4
4*(V[n])²+(n+1)^3)=n^4+2n^3+n²+4n^3+12n²+12n+4
=n^4+6n^3+13n²+12n+4
=n^4+2n^3+4n^3+8n²+5n²+10n+2n+4
=n^3(n+2)+4n²(n+2)+5n(n+2)+2(n+2)
=(n+2)(n^3+4n²+5n+2)=(n+2)(n^3+2n^2+2n^2+4n+n+2)
=(n+2)[n²(n+2)+2n(n+2)+1(n+2)]
=(n+2)(n+2)(n²+2n+1)
=(n+1)²*(n+2)²
=> (V[n])²+(n+1)²={(n+1)/n+2)/2}²={V[n+1]}²
W[n+1]={V[n+1]}²