Question

Sa se calcuzele: 1^3 + 3^3 + 5^3 + …+(2n-1)^3
Rezultatul trebuie sa fie n^2(2n^2 -1)
Ofer coroana!

Answer 1

Răspuns:

Explicație pas cu pas:

folosim formula:

$1^3 + 2^3 + ... + n^3 = [\frac{n(n+1)}{2}]^2$

facem un mic artificiu de calcul:

$1^3 +3^3 + 5^3 + ... + (2n-1)^3 = 1^3 +2^3 - 2^3 +3^3 + 4^3 - 4^3 + 5^3 + ... + (2n-2)^3 - (2n-2)^3 +(2n-1)^3 =$

$= 1^3 +2^3 +3^3 + 4^3+ ... +(2n-1)^3 - 2^3 - 4^3 - ... -(2n-2)^3 =$

$= 1^3 +2^3 +3^3 + 4^3+ ... +(2n-1)^3 - (2*1)^3 - (2*2)^3 - ... -[2*(n-1)]^3 =$

$= 1^3 +2^3 +3^3 + 4^3+ ... + (2n-1)^3 - 2^3*1^3 - 2^3*2^3 - ... -2^3*(n-1)^3 =$

$= 1^3 +2^3 +3^3 + 4^3+ ... + (2n-1)^3 - 2^3*[1^3 +2^3 + ... + (n-1)^3] =$

$= [\frac{(2n-1)(2n-1+1)}{2}]^2 -2^3* [\frac{(n-1)(n-1+1)}{2} ]^2 =$

$= [\frac{2n*(2n-1)}{2}]^2 -8*[\frac{n(n-1)}{2}]^2 =$

$= \frac{4n^2*(2n-1)^2}{4} -\frac{8n^2(n-1)^2}{4} =$

$= \frac{4n^2*(4n^2 - 4n + 1)-8n^2(n^2 - 2n + 1)}{4} =$

$= \frac{16n^4- 16n^3 + 4n^2 -8n^4 + 16n^3 - 8n^2}{4} =$

$= \frac{8n^4-4n^2}{4} =$

$= \frac{4n^2(2n^2-1)}{4} =$

$= n^2(2n^2-1)$