Question

a)aratati ca
$\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n \times (n + 1)}$
pentru orice n->numar natural
b)calculati:
$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ... + \frac{1}{2013 \times 2014}$
c)aratati ca (poza) pentru orice n,m ->numar natural
p.s:la punctul c exercitiul e in atasare​

Answer 1

a)

$\frac{1}{n} - \frac{1}{n+1} = \frac{n+1}{n(n+1)} - \frac{n}{n(n+1)} = \frac{n+1-n}{n(n+1)} = \frac{1}{n(n+1)}$

b)

$\textrm{Folosind ce am calculat la punctul a, avem:}\\\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots + \frac{1}{2013\cdot 2014} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{2013} + \frac{1}{2013} - \frac{1}{2014}\\\\\textrm{Acum se reduc si ramanem doar cu:}\\\frac{1}{1} - \frac{1}{2014} = 1 - \frac{1}{2014} = \frac{2014 - 1}{2014} = \frac{2013}{2014}$

c)

$\frac{1}{n} - \frac{1}{n+m} = \frac{m}{n\cdot(n+m)}\\\\\textrm{Incepem cu partea stanga a egalitatii:}\\\frac{1}{n} - \frac{1}{n+m} = \frac{n+m}{n(n+m)} - \frac{n}{n(n+m)} = \frac{n+m - n}{n(n+m)} = \frac{m}{n\cdot(n+m)}$

Answer 2

Explicație pas cu pas:

a)

1/n-1/(n+1)=(n+1-n)/n(n+1)=1/n(n+1)

b)

1/(1*2)+1/(2*3)+...+1/(2013*2014)

=(2-1)/(1*2)+(3-2)/(2*3)+...+(2014-2013)/(2014*2013)

=2/(1*2)-1/(1*2)+3(2*3)-2/(2*3)+...+2014/(2014*2013)-2013/(2014*2013)

=1-1/2+1/2-1/3+...+1/2013-1/2014=>Efectuam reducerea termenilor asemenea

adica -1/2+1/2, -1/3+1/3 etc..

avem ca suma este egala cu 1-1/2014=(2014-1)/2014=2013/2014

La ce tot trebuie sa arati cred...

1/n-1/(n+m)=(n+m-n(/n(n+m)=m/n(n+m)

Bafta!