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Appliquer la formule du binôme de Newton - Exercice 1

12 min
25
Utiliser la formule du binôme pour développer :
Question 1

(1+z)3\left(1+z\right)^{3}

Correction
    Formule du binoˆme de Newton\red{\text{Formule du binôme de Newton}}
Soient aa et bb deux nombres complexes. Pour tout entier naturel nn, on a : (a+b)n=∑k=0n(nk)akbn−k\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
(1+z)3=∑k=0n(3k)1kz3−k\left(1+z\right)^{3} =\sum _{k=0}^{n}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) 1^{k} z^{3-k} équivaut successivement à :
(1+z)3=(30)10z3−0+(31)11z3−1+(32)12z3−2+(32)13z3−3\left(1+z\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)1^{0} z^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)1^{1} z^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)1^{2} z^{3-2} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)1^{3} z^{3-3}
(1+z)3=(30)10z3+(31)11z2+(32)12z1+(33)13z0\left(1+z\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}1^{0} z^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}1^{1} z^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}1^{2} z^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}1^{3} z^{0}
(1+z)3=1×10×z3+3×11×z2+3×12×z1+1×13×z0\left(1+z\right)^{3} =\red{1}\times 1^{0} \times z^{3} +\blue{3}\times 1^{1} \times z^{2} +\pink{3}\times 1^{2} \times z^{1} +\purple{1}\times 1^{3} \times z^{0}
(1+z)3=1×1×z3+3×1×z2+3×1×z1+1×1×1\left(1+z\right)^{3} =1\times 1\times z^{3} +3\times 1\times z^{2} +3\times 1\times z^{1} +1\times 1\times 1
Ainsi :
(1+z)3=z3+3z2+3z+1\left(1+z\right)^{3} =z^{3} +3z^{2} +3z+1

Question 2

(2+z)4\left(2+z\right)^{4}

Correction
    Formule du binoˆme de Newton\red{\text{Formule du binôme de Newton}}
Soient aa et bb deux nombres complexes. Pour tout entier naturel nn, on a : (a+b)n=∑k=0n(nk)akbn−k\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
(2+z)4=∑k=0n(4k)2kz4−k\left(2+z\right)^{4} =\sum _{k=0}^{n}\left(\begin{array}{c} {4} \\ {k} \end{array}\right) 2^{k} z^{4-k} équivaut successivement à :
(2+z)4=(40)20z4+(41)21z4−1+(42)22z4−2+(43)23z4−3+(44)24z4−4\left(2+z\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)2^{0} z^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)2^{1} z^{4-1} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)2^{2} z^{4-2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)2^{3} z^{4-3} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)2^{4} z^{4-4}
(2+z)4=(40)20z4+(41)21z3+(42)22z2+(43)23z1+(44)24z0\left(2+z\right)^{4} =\left(\begin{array}{c} {4} \\ {0} \end{array}\right)2^{0} z^{4} +\left(\begin{array}{c} {4} \\ {1} \end{array}\right)2^{1} z^{3} +\left(\begin{array}{c} {4} \\ {2} \end{array}\right)2^{2} z^{2} +\left(\begin{array}{c} {4} \\ {3} \end{array}\right)2^{3} z^{1} +\left(\begin{array}{c} {4} \\ {4} \end{array}\right)2^{4} z^{0}
(2+z)4=(40)20z4+(41)21z3+(42)22z2+(43)23z1+(44)24z0\left(2+z\right)^{4} =\red{\left(\begin{array}{c} {4} \\ {0} \end{array}\right)}2^{0} z^{4} +\blue{\left(\begin{array}{c} {4} \\ {1} \end{array}\right)}2^{1} z^{3} +\pink{\left(\begin{array}{c} {4} \\ {2} \end{array}\right)}2^{2} z^{2} +\purple{\left(\begin{array}{c} {4} \\ {3} \end{array}\right)}2^{3} z^{1} +\green{\left(\begin{array}{c} {4} \\ {4} \end{array}\right)}2^{4} z^{0}
(2+z)4=1×20z4+4×21z3+6×22z2+4×23z1+1×24z0\left(2+z\right)^{4} =\red{1}\times 2^{0} z^{4} +\blue{4}\times 2^{1} z^{3} +\pink{6}\times 2^{2} z^{2} +\purple{4}\times 2^{3} z^{1} +\green{1}\times 2^{4} z^{0}
(2+z)4=1×1×z4+4×2×z3+6×4×z2+4×8×z1+1×16×z0\left(2+z\right)^{4} =1\times 1\times z^{4} +4\times 2\times z^{3} +6\times 4\times z^{2} +4\times 8\times z^{1} +1\times 16\times z^{0}
Ainsi :
(2+z)4=z4+8z3+24z2+32z+16\left(2+z\right)^{4} =z^{4} +8z^{3} +24z^{2} +32z +16

Question 3

(z−3)3\left(z-3\right)^{3}

Correction
    Formule du binoˆme de Newton\red{\text{Formule du binôme de Newton}}
Soient aa et bb deux nombres complexes. Pour tout entier naturel nn, on a : (a+b)n=∑k=0n(nk)akbn−k\left(a+b\right)^{n} =\sum _{k=0}^{n}\left(\begin{array}{c} {n} \\ {k} \end{array}\right) a^{k} b^{n-k}
(z−3)3=∑k=03(3k)zk(−3)3−k\left(z-3\right)^{3} =\sum _{k=0}^{3}\left(\begin{array}{c} {3} \\ {k} \end{array}\right) z^{k} \left(-3\right)^{3-k} équivaut successivement à :
(z−3)3=(30)z0(−3)3−0+(31)z1(−3)3−1+(32)z2(−3)3−2+(33)z3(−3)3−3\left(z-3\right)^{3} =\left(\begin{array}{c} {3} \\ {0} \end{array}\right)z^{0} \left(-3\right)^{3-0} +\left(\begin{array}{c} {3} \\ {1} \end{array}\right)z^{1} \left(-3\right)^{3-1} +\left(\begin{array}{c} {3} \\ {2} \end{array}\right)z^{2} \left(-3\right)^{3-2} +\left(\begin{array}{c} {3} \\ {3} \end{array}\right)z^{3} \left(-3\right)^{3-3}
(z−3)3=(30)z0(−3)3+(31)z1(−3)2+(32)z2(−3)1+(33)z3(−3)0\left(z-3\right)^{3} =\red{\left(\begin{array}{c} {3} \\ {0} \end{array}\right)}z^{0} \left(-3\right)^{3} +\blue{\left(\begin{array}{c} {3} \\ {1} \end{array}\right)}z^{1} \left(-3\right)^{2} +\pink{\left(\begin{array}{c} {3} \\ {2} \end{array}\right)}z^{2} \left(-3\right)^{1} +\purple{\left(\begin{array}{c} {3} \\ {3} \end{array}\right)}z^{3} \left(-3\right)^{0}
(z−3)3=1×z0(−3)3+3×z1(−3)2+3×z2(−3)1+1×z3(−3)0\left(z-3\right)^{3} =\red{1}\times z^{0} \left(-3\right)^{3} +\blue{3}\times z^{1} \left(-3\right)^{2} +\pink{3}\times z^{2} \left(-3\right)^{1} +\purple{1}\times z^{3} \left(-3\right)^{0}
(z−3)3=1×z0(−3)3+3×z1(−3)2+3×z2(−3)1+1×z3(−3)0\left(z-3\right)^{3} =1\times z^{0} \left(-3\right)^{3} +3\times z^{1} \left(-3\right)^{2} +3\times z^{2} \left(-3\right)^{1} +1\times z^{3} \left(-3\right)^{0}
(z−3)3=1×1×(−27)+3×z×9+3×z2×(−3)+1×z3×1\left(z-3\right)^{3} =1\times 1\times \left(-27\right)+3\times z\times 9+3\times z^{2} \times \left(-3\right)+1\times z^{3} \times 1
(z−3)3=−27+27z−9z2+z3\left(z-3\right)^{3} =-27+27z-9z^{2} +z^{3}
Ainsi :
(z−3)3=z3−9z2+27z−27\left(z-3\right)^{3} =z^{3} -9z^{2} +27z-27