Question

Sa se demonstreze ca pentru n e N au loc egalitatile
a)
$\frac{1}{1 \times2 } + \frac{1}{2 \times 3} + ... + \frac{n}{n(n + 1)} = \frac{n}{n + 1}$
b)
$\frac{1}{3 \times 5} + \frac{1}{3 \times 5} + ... + \frac{1}{(2n - 1)(2n + 1) } = \frac{n}{2n + 1}$
c)
$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + ... + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}$
Urgeeeent va rooog!​

Answer 1

Răspuns:

a)

[tex]\frac{1}{1*2} = \frac{1}{1} - \frac{1}{2}\\ \frac{1}{2*3} = \frac{1}{2} - \frac{1}{3}\\ .....\\ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}[/tex]

⇒

[tex]\frac{1}{1*2} + \frac{1}{2*3} +.....+\frac{1}{n(n+1)} =\\ = \frac{1}{1} - \frac{1}{2}+ \frac{1}{2} - \frac{1}{3}+....+ \frac{1}{n} - \frac{1}{n+1} =\\ = \frac{1}{1} -\frac{1}{n+1} =\\ = \frac{n+1-1}{n+1}=\\ = \frac{n}{n+1}[/tex]

b)

$\frac{1}{(2k-1)(2k+1)} = \frac{1}{2} [\frac{1}{2k-1} -\frac{1}{2k+1} ]$

⇒

[tex]\frac{1}{1*3} +\frac{1}{3*5} +...+\frac{1}{(2n-1)(2n+1)} =\\ = \frac{1}{2} [\frac{1}{1} -\frac{1}{3} ]+\frac{1}{2} [\frac{1}{3} -\frac{1}{5} ]+...+\frac{1}{2} [\frac{1}{2n-1} -\frac{1}{2n+1} ]=\\ = \frac{1}{2} [\frac{1}{1} -\frac{1}{3} +\frac{1}{3} -\frac{1}{5} +...+\frac{1}{2n-1} -\frac{1}{2n+1} ]=\\ = \frac{1}{2} [\frac{1}{1} -\frac{1}{2n+1} ]=\\ = \frac{1}{2}\frac{2n+1-1}{2n+1} =\\ = \frac{1}{2}\frac{2n}{2n+1} =\\ = \frac{n}{2n+1} [/tex]

c)

[tex]\frac{1}{(3k-2)(3k+1)} = \frac{1}{3} [\frac{1}{3k-2} -\frac{1}{3k+1} ]\\ [/tex]

⇒

[tex]\frac{1}{3} [\frac{1}{1} -\frac{1}{4} ]+\frac{1}{3} [\frac{1}{4} -\frac{1}{7} ]+...+ \frac{1}{3} [\frac{1}{3n-2} -\frac{1}{3n+1} ] =\\ = \frac{1}{3} [\frac{1}{1} -\frac{1}{4} +\frac{1}{4} -\frac{1}{7} +...+\frac{1}{3n-2} -\frac{1}{3n+1} ] =\\ = \frac{1}{3} [\frac{1}{1} -\frac{1}{3n+1} ] =\\ = \frac{1}{3} [\frac{3n+1-1}{3n+1} ] =\\ = \frac{1}{3} [\frac{3n}{3n+1} ] =\\ = \frac{n}{3n+1} [/tex]