Răspuns:
Explicație pas cu pas:
folosim formula:
[tex]1^3 + 2^3 + ... + n^3 = [\frac{n(n+1)}{2}]^2[/tex]
facem un mic artificiu de calcul:
[tex]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 =[/tex]
[tex]= 1^3 +2^3 +3^3 + 4^3+ ... +(2n-1)^3 - 2^3 - 4^3 - ... -(2n-2)^3 =[/tex]
[tex]= 1^3 +2^3 +3^3 + 4^3+ ... +(2n-1)^3 - (2*1)^3 - (2*2)^3 - ... -[2*(n-1)]^3 =[/tex]
[tex]= 1^3 +2^3 +3^3 + 4^3+ ... + (2n-1)^3 - 2^3*1^3 - 2^3*2^3 - ... -2^3*(n-1)^3 =[/tex]
[tex]= 1^3 +2^3 +3^3 + 4^3+ ... + (2n-1)^3 - 2^3*[1^3 +2^3 + ... + (n-1)^3] =[/tex]
[tex]= [\frac{(2n-1)(2n-1+1)}{2}]^2 -2^3* [\frac{(n-1)(n-1+1)}{2} ]^2 =[/tex]
[tex]= [\frac{2n*(2n-1)}{2}]^2 -8*[\frac{n(n-1)}{2}]^2 =[/tex]
[tex]= \frac{4n^2*(2n-1)^2}{4} -\frac{8n^2(n-1)^2}{4} =[/tex]
[tex]= \frac{4n^2*(4n^2 - 4n + 1)-8n^2(n^2 - 2n + 1)}{4} =[/tex]
[tex]= \frac{16n^4- 16n^3 + 4n^2 -8n^4 + 16n^3 - 8n^2}{4} =[/tex]
[tex]= \frac{8n^4-4n^2}{4} =[/tex]
[tex]= \frac{4n^2(2n^2-1)}{4} =[/tex]
[tex]= n^2(2n^2-1)[/tex]
