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Exercice 17

Basé sur la formule du binôme de Newton :

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

Somme ( $\sum$ )	$a \cdot \binom{n}{k} x^{n-k} y^k$	$a, x, y$	Réponse finale
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a) $\sum_{k=0}^{n} \binom{n}{k} 2^k 3^{n-k}$	$\sum \binom{n}{k} 3^{n-k} 2^k$	$a = 1, \, x = 3, \, y = 2$	$5^n$
b) $\sum_{k=0}^{n} \binom{n}{k} 2^k \left(\frac{1}{2}\right)^{n-k}$	$\sum \binom{n}{k} \left(\frac{1}{2}\right)^{n-k} 2^k$	$a = 1, \, x = \frac{1}{2}, \, y = 2$	$\left(\frac{5}{2}\right)^n$
c) $\sum_{k=0}^{n} \binom{n}{k} 3^{k+1} 5^{n-k}$	$\sum 3 \cdot \binom{n}{k} 5^{n-k} 3^k$	$a = 3, \, x = 5, \, y = 3$	$3 \cdot 8^n$
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d) $\sum_{k=0}^{n} \binom{n}{k} 2^{k+1} 3^{2n-k}$	$\sum 2 \cdot \binom{n}{k} (3^2)^{n-k} 2^k$	$a = 2, \, x = 9, \, y = 2$	$2 \cdot 11^n$
e) $\sum_{k=0}^{n} \binom{n}{k} 2^k$	$\sum \binom{n}{k} 1^{n-k} 2^k$	$a = 1, \, x = 1, \, y = 2$	$3^n$
f) $\sum_{k=0}^{n} \binom{n}{k} 4^k 3^{n-k}$	$\sum \binom{n}{k} 3^{n-k} 4^k$	$a = 1, \, x = 3, \, y = 4$	$7^n$
g) $\sum_{k=0}^{n} \binom{n}{k} \frac{5^k}{2^{n-k}}$	$\sum \binom{n}{k} \left(\frac{1}{2}\right)^{n-k} 5^k$	$a = 1, \, x = \frac{1}{2}, \, y = 5$	$\left(\frac{11}{2}\right)^n$
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h) $\sum_{k=0}^{n} \binom{n}{k} 3^{2k-n}$	$\sum \frac{1}{3^n} \binom{n}{k} 1^{n-k} (9^k)$	$a = \frac{1}{3^n}, \, x = 1, \, y = 9$	$\left(\frac{10}{3}\right)^n$
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i) $\sum_{k=1}^{n} \binom{n}{k} 5^k 3^{n-k}$	Somme de $k=1$ à $n$ (Retirer le terme $k=0$ )	$x = 3, \, y = 5$	$8^n - 3^n$
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j) $\sum_{k=0}^{n-1} \binom{n}{k} 3^k 4^{n-k}$	Somme de $0$ à $n-1$ (Retirer le terme $k=n$ )	$x = 4, \, y = 3$	$7^n - 3^n$