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17. Suma 222 2 222 2 3 1 5 1 7 1 101 1 ... 3 1 5 1 7 1 101 1 S         are valoarea: a. 2575 51 ; b. 2601 51 ; c. 101 2 ; d. 51.​

17 Suma 222 2 222 2 3 1 5 1 7 1 101 1 3 1 5 1 7 1 101 1 S Are Valoarea A 2575 51 B 2601 51 C 101 2 D 51 class=

Răspuns :

Răspuns:

Explicație pas cu pas:

Vezi imaginea Stefanboiu

Răspuns:

[tex]\boldsymbol{ \red{ a. \ \dfrac{2575}{51}}}[/tex]

Explicație pas cu pas:

[tex]S = \dfrac{3^2 - 1 + 2}{3^2 - 1} + \dfrac{5^2 - 1 + 2}{5^2 - 1} + \dfrac{7^2 - 1 + 2}{7^2 - 1} + ... + \dfrac{101^2 - 1 + 2}{101^2 - 1}\\[/tex]

[tex]S = 1 + \dfrac{2}{(3-1)(3+1)} + 1 + \dfrac{2}{(5-1)(5+1)} + 1 + \dfrac{2}{(7 - 1)(7 + 1)} + ... + 1 + \dfrac{2}{(101-1)(101+1)}\\[/tex]

(101 - 3) : 2 + 1 = 98 : 2 + 1 = 49 + 1 = 50 (termeni)

[tex]S = \underbrace{1 + 1 + 1 + ... + 1}_{50} + \dfrac{2}{2 \cdot 4} + \dfrac{2}{4 \cdot 6} + \dfrac{2}{6 \cdot 8} + ... + \dfrac{2}{100 \cdot 102}\\[/tex]

[tex]S = 50 + \dfrac{4 - 22}{2 \cdot 4} + \dfrac{6 - 4}{4 \cdot 6} + \dfrac{8 - 6}{6 \cdot 8} + ... + \dfrac{102 - 100}{100 \cdot 102}\\[/tex]

[tex]S = 50 + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{6} + \dfrac{1}{6} - \dfrac{1}{8} + ... + \dfrac{1}{100} - \dfrac{1}{102} \\[/tex]

[tex]S = 50 + \dfrac{1}{2} - \dfrac{1}{102} = \dfrac{50 \cdot 102 + 51 - 1}{102} = \dfrac{5150}{102}^{(2}\\[/tex]

[tex]\bf S = \dfrac{2575}{51}[/tex]