Réponse :
n
∑ i = 3 + 4 + 5 + ..... + n = (3+n)(n-3+1)/2 = (3+n)(n-2)/2
i = 3
∑ (2i - 1) = 1 + 3 + 5 + .....+ (2n - 1) = (1+2n-1)n/2 = n²
i = 1
n+1
∑ (3k+7) = 19 + 22 + 25 + ..... + (3(n+1) + 7)(n - 17)/2 = (3n + 10)(n - 17)/2
i = 4
Explications étape par étape :
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Réponse :
n
∑ i = 3 + 4 + 5 + ..... + n = (3+n)(n-3+1)/2 = (3+n)(n-2)/2
i = 3
n
∑ (2i - 1) = 1 + 3 + 5 + .....+ (2n - 1) = (1+2n-1)n/2 = n²
i = 1
n+1
∑ (3k+7) = 19 + 22 + 25 + ..... + (3(n+1) + 7)(n - 17)/2 = (3n + 10)(n - 17)/2
i = 4
Explications étape par étape :