Răspuns:
Explicație pas cu pas:
Notăm
[tex]P(n) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2n-1} - \dfrac{1}{2n} = \dfrac{1}{n+1} + \dfrac{1}{n+2} + ... +\dfrac{1}{2n} \\[/tex]
[tex]P(n) \ \ \dfrac{1}{1} - \dfrac{1}{2} = \dfrac{1}{1+1} \iff \dfrac{1}{2} = \dfrac{1}{2} \to (A)\\[/tex]
[tex]P(k) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k} = \dfrac{1}{k+1} + \dfrac{1}{k+2} + ... +\dfrac{1}{2k} \to (A) \\[/tex]
și vom demonstra că P(k+1) este adevărată, unde k ≥ 1
[tex]P(k + 1) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k} + \dfrac{1}{2k+1} - \dfrac{1}{2k+2} = \\[/tex]
[tex]= \underbrace{ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k}}_{P(k)} + \dfrac{1}{2k+1} - \dfrac{1}{2k+2} \\[/tex]
[tex]= {\bf\dfrac{1}{k+1}} + \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} - {\bf\dfrac{1}{2k+2}}\\[/tex]
[tex]= \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} + \bigg(\dfrac{1}{k+1} - \dfrac{1}{2k+2}\bigg)[/tex]
[tex]= \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} + \bigg(\dfrac{2}{2(k+1)} - \dfrac{1}{2(k+1)}\bigg)[/tex]
[tex]= \dfrac{1}{(k+1)+1} + ... +\dfrac{1}{2(k+1)-2} + \dfrac{1}{2(k+1)-1} + \dfrac{1}{2(k+1)} \\[/tex]
[tex]\implies P(k + 1) \to adev\breve{a}rat\breve{a} \implies \boldsymbol{P(n) \ este \ adev\breve{a}rat\breve{a} \ \forall n \in \Bbb{N^{\ast}}}\\[/tex]
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Răspuns:
Explicație pas cu pas:
Demonstrație prin inducție matematică
Notăm
[tex]P(n) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2n-1} - \dfrac{1}{2n} = \dfrac{1}{n+1} + \dfrac{1}{n+2} + ... +\dfrac{1}{2n} \\[/tex]
[tex]P(n) \ \ \dfrac{1}{1} - \dfrac{1}{2} = \dfrac{1}{1+1} \iff \dfrac{1}{2} = \dfrac{1}{2} \to (A)\\[/tex]
[tex]P(k) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k} = \dfrac{1}{k+1} + \dfrac{1}{k+2} + ... +\dfrac{1}{2k} \to (A) \\[/tex]
și vom demonstra că P(k+1) este adevărată, unde k ≥ 1
[tex]P(k + 1) \ \ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k} + \dfrac{1}{2k+1} - \dfrac{1}{2k+2} = \\[/tex]
[tex]= \underbrace{ \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{2k-1} - \dfrac{1}{2k}}_{P(k)} + \dfrac{1}{2k+1} - \dfrac{1}{2k+2} \\[/tex]
[tex]= {\bf\dfrac{1}{k+1}} + \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} - {\bf\dfrac{1}{2k+2}}\\[/tex]
[tex]= \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} + \bigg(\dfrac{1}{k+1} - \dfrac{1}{2k+2}\bigg)[/tex]
[tex]= \dfrac{1}{k+2} + ... +\dfrac{1}{2k} + \dfrac{1}{2k+1} + \bigg(\dfrac{2}{2(k+1)} - \dfrac{1}{2(k+1)}\bigg)[/tex]
[tex]= \dfrac{1}{(k+1)+1} + ... +\dfrac{1}{2(k+1)-2} + \dfrac{1}{2(k+1)-1} + \dfrac{1}{2(k+1)} \\[/tex]
[tex]\implies P(k + 1) \to adev\breve{a}rat\breve{a} \implies \boldsymbol{P(n) \ este \ adev\breve{a}rat\breve{a} \ \forall n \in \Bbb{N^{\ast}}}\\[/tex]