Question

Poate cineva să mă ajute?

Answer 1

Răspuns:

Explicație pas cu pas:

$1.~Verificam~ pentru~n=1~ \frac{1*(3*1-1)}{2}=\frac{1*2}{2}=1,~adevarat.\\2.~Admitem~ca~e~adevarata~pt.~n=k,~adica~1+4+7+...+(3k-2)=\frac{k(3k-1)}{2}.\\3.~verificam~adevarul~pt.~n=k+1,~adica~tr.~sa~obtinem~1+4+7+...+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}.\\1+4+7+...+(3k-2)+(3(k+1)-2)=\frac{k(3k-1)}{2}+(3(k+1)-2)=\frac{k(3k-1)}{2}+(3k+1)=\frac{k(3k-1)+2(3k+1)}{2}=\frac{3k^{2}+5k+2}{2}=\frac{3k^{2}+6k+3-k-1}{2}=\frac{3(k^{2}+2k+1)-(k+1)}{2}=\frac{3(k+1)^{2}-(k+1)}{2}=\frac{(k+1)(3(k+1)-1)}{2}$

am obtinut la ce am vrut sa ajungem, deci egalitatea este adevarata pentru ∀n∈N, n>0.