Question

Se consideră suma :

S= 1/1•3+1/3•5+1/5•7+....+1/(2n-1)•(2n+1)

Aflați n astfel ca S= 7/15

Answer 1

Mai intai vom gasi un mod de a calcula mai usor suma.

O sa scriem termenii in felul urmator:

[tex] \frac{1}{1*3}= \frac{1}{2} ( \frac{1}{1} - \frac{1}{3} ) \\ \frac{1}{3*5}= \frac{1}{2} ( \frac{1}{3} - \frac{1}{5} ) \\ \frac{1}{5*7} = \frac{1}{2} ( \frac{1}{5} - \frac{1}{7} )[/tex]

Si tot asa, pana cand la ultimul element:
$\frac{1}{(2n-1)(2n+1)}= \frac{1}{2} ( \frac{1}{2n-1} - \frac{1}{2n+1} )$


Se observa ca toti termenii il au ca factor comun pe 1/2:

$S= \frac{1}{2}( \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} +...+ \frac{1}{2n-1} - \frac{1}{2n+1} )$

Se reduc toti termenii din paranteza, mai putin primul si ultimul:

$S= \frac{1}{2} ( \frac{1}{1} - \frac{1}{2n+1} )= \frac{1}{2} * \frac{2n}{2n+1}= \frac{n}{2n+1}$

Acum, trebuie sa-l aflam pe n, astfel incat n/(2n+1) = 7/15 ==> n = 7

Answer 2

Utlizam formula
1/ n(n+k)=1/k(1/n -1/ n+k)
S=1/1·3 +1/3·5 +1/5·7 +1/(2n-1)(2n+1)
1/1·3=1/2(1-1/3)
1/3·5=1/2(1/3-1/5)
......
1/(2n-1)(2n+1)=1/2(1/2n-1 -1/2n+1)
Insuman termen cu termen obtinem ca
S=1/2(1-1/2n+1)=1/2 ·2n /2n+1
Dar deoarece S=7/15 ⇒1/2 ·2n/2n+1=7/15 ⇒30n=7·2(2n+1) ⇒30n=28n+14 ⇒n=7 .