Question

Va rog sa ma ajutati....1 a b c

Answer 1

a) 1+2+3+..+n=(n(n+1))/2

I.verific :
pt n=1 =>(1(1+1))/2 = 1 => 2/2 = 1 adv
pt n=2 =>(2(2+1))/2= 3 => 3 = 3 adv ...

||. demonstrez ca P(k) -> P(k+1) adv
P(k)=1+2+...+k=(k(k+1))/2
P(k+1)=1+2+...+k+(k+1)=(k+1(k+1+1))/2 = ((k+1)(k+2))/2 =(k^2+3k+2)/2

=>1+2+3+...+k+k+1 =((k^2+3k+2)/2
(k(k+1))/2 + k+1 =((k^2+3k+2)/2
(k^2+k+2k+2)/2=((k^2+3k+2)/2
((k^2+3k+2)/2=((k^2+3k+2)/2

Din I si II => P(k+1) adv => propozitia este adv

b) 1^2+2^2+..+n^2=((n(n+1)(2n+1))/6

I : pt n=1 => (1(1+1)(2+1))/6=1 => 6/6=1 adv ...

II: P(k) -> P(k+1) adv
P(k)=1^2+2^2+...+k^2=(k(k+1)(2k+1))/6
P(k+1)=1^2+2^2+...+k^2+(k+1)^2=((k+1)(k+2)(2(k+1)+1))/6=((k^2+3k+2)(2k+3))/6=(k^3+8k^2+7k+6)/6
1^2+2^2+...+k^2+(k+1)^2 = (k^3+8k^2+7k+6)/6
(k(k+1)(2k+1))/6+(k+1)^2 = (k^3+8k^2+7k+6)/6
(k^3+8k^2+7k+6)/6=(k^3+8k^2+7k+6)/6
(Consider ca poti face si tu calculele acestea simple pt a arata egalitatea )
Din I si II => P(k+1) adv => propozitia este adv

c) Ti l las pe acesta sa l rezolvi individual. In plus, se calculeza exact ca la punctele anterioare expuse mai sus .Dar in caz daca ai vreo nelamuriere legat de acest subpunct , imi poti lasa un comentariu
Bafta !