Răspuns:
Explicație pas cu pas:
g)
[tex]Notam \ \ \Big{S _n} = 1 \cdot2 + 2 \cdot3 + ... + n(n + 1) = \frac{\Big{n(n+1)(n+2)}}{\Big3}[/tex]
[tex]\Big{S _{n+1}} = 1 \cdot2 + 2 \cdot3 + ... + n (n + 1)+ (n+1)(n + 2) =[/tex]
[tex]=\frac{\Big{n(n+1)(n+2)}}{\Big3} + (n+1)(n + 2) = \frac{\Big{n(n+1)(n+2) + 3(n+1)(n + 2)}}{\Big3} =[/tex]
[tex]\frac{\Big{ \ [(n+1)(n+2)](n + 3)}}{\Big3} = \frac{\Big{ \ (n+1)(n+2)(n + 3)}}{\Big3}[/tex]
h)
[tex]Notam \ \ \Big{S _n} = 1 \cdot2\cdot3 + 2 \cdot3\cdot4 + ... + n(n + 1)(n+2) = \frac{\Big{n(n+1)(n+2)(n+2)}}{\Big4}[/tex]
[tex]\Big{S _{n+1}} = 1 \cdot2\cdot3 + 2 \cdot3\cdot4 + ... + n (n + 1)(n+2) +(n+1)(n + 2)(n+3) =[/tex]
[tex]= \frac{\Big{n(n+1)(n+2)(n+2)}}{\Big4} +(n+1)(n + 2)(n+3) =[/tex]
[tex]= \frac{\Big{n(n+1)(n+2)(n+3) +4(n+1)(n + 2)(n+3) }}{\Big4} =[/tex]
[tex]= \frac{\Big{\ [(n+1)(n+2)(n+3)](n +4) }}{\Big4} = \frac{\Big{\ (n+1)(n+2)(n+3)(n +4) }}{\Big4}[/tex]
i)
[tex]Notam \ \ \Big{S _n} = 1 \cdot4 + 2 \cdot7 + 3 \cdot10 + ... + n(3n + 1) = n(n+1)^2[/tex]
[tex]\Big{S _{n+1}} = 1 \cdot4 + 2 \cdot7 + 3 \cdot10 + ... + n(3n + 1) + (n+1)[3(n+1)+1] =[/tex]
[tex]= n(n+1)^2 + (n+1)(3n+4) = (n+1)[n(n+1) + (3n + 4)] =[/tex]
[tex]= (n+1)(n^2+n + 3n +4 ) = (n+1)(n^2+ 4n +4 ) =(n+1)(n+2)^2[/tex]
j)
[tex]Notam \ \ \Big{S _n} = 1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{n+1}n^2 = (-1)^{n+1}\frac{\Big{n(n+1)}}{\Big2}[/tex]
[tex]\Big{S _{n+1}} = 1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{n+1}n^2 + (-1)^{(n+1)+1}(n+1)^2=[/tex]
[tex]= (-1)^{n+1}\frac{\Big{n(n+1)}}{\Big2}+ (-1)^{(n+1)+1}(n+1)^2 =[/tex]
[tex]= (-1)^{n+1}\frac{\Big{n(n+1)}}{\Big2}+ (-1)^{(n+1)}(-1)(n+1)^2=[/tex]
[tex]= (-1)^{n+1}\Big[\frac{\Big{n(n+1)}}{\Big2} -(n+1)^2\Big] = (-1)^{n+1}\frac{\Big{n(n+1)-2(n+1)^2}}{\Big2}=[/tex]
[tex]= (-1)^{n+1}\frac{\Big{\ [(n+1)[n-2(n+1)]}}{\Big2}= (-1)^{n+1}\frac{\Big{\ (n+1)(n-2n-2)}}{\Big2}=[/tex]
[tex]= (-1)^{n+1}\frac{\Big{\ (n+1)(-n - 2)}}{\Big2}= (-1)^{n+1}(-1)\frac{\Big{\ (n+1)(n + 2)}}{\Big2}=[/tex]
[tex](-1)^{n+2}\frac{\Big{\ (n+1)(n + 2)}}{\Big2}[/tex]
