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a fost răspuns

Efectuati: (1-1/2)+(1-5/6)+(1-11/12)+...+(1-109/110).

Răspuns :

[tex](1-\frac{1}{2} ) + (1 - \frac{5}{6}) + (1 - \frac{1}{12}) + ... + (1 - \frac{109}{110})=[/tex]

[tex]\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{110} = \frac{1}{2*1} + \frac{1}{3*2} + \frac{1}{4*3} + .... + \frac{1}{11*10} =[/tex]

[tex]\frac{2-1}{2*1} + \frac{3-2}{3*2} + \frac{4-3}{4*3} + ... \frac{11-10}{11*10} =[/tex]

[tex]\frac{2}{2*1} - \frac{1}{2*1} + \frac{3}{3*2} - \frac{2}{3*2} + \frac{4}{4*3} - \frac{3}{4*3} + ... + \frac{11}{11*10} - \frac{10}{11*10}=[/tex]

[tex]1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - ... - \frac{1}{10} + \frac{1}{10} - \frac{1}{11} =[/tex]

[tex]1- \frac{1}{10} = \frac{10}{12}[/tex]

* Înseamnă înmulțire .

Sper că te-am ajutat!

Succes!

[tex]\bf \Big(1-\dfrac{1}{2}\Big)+\Big(1-\dfrac{5}{6}\Big)+\Big(1-\dfrac{1}{12}\Big)+....+\Big(1-\dfrac{109}{110}\Big)=[/tex]

[tex]\bf \Big(\dfrac{1\cdot2}{2}-\dfrac{1}{2}\Big)+\Big(\dfrac{1\cdot6}{6}-\dfrac{5}{6}\Big)+\Big(\dfrac{1\cdot12}{12}-\dfrac{1}{12}\Big)+....+\Big(\dfrac{1\cdot110}{110}-\dfrac{109}{110}\Big)=[/tex]

[tex]\bf \dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+....+\dfrac{1}{110}=[/tex]

[tex]\bf \dfrac{1}{2\cdot 1}+\dfrac{1}{3\cdot 2}+\dfrac{1}{4\cdot 3}+....+\dfrac{1}{11\cdot 10}=[/tex]

incercam sa il scriem pe 1 ca: 2-1 sau 3-2 ori 4-3.......

[tex]\bf \dfrac{2-1}{2\cdot 1}+\dfrac{3-2}{3\cdot 2}+\dfrac{4-3}{4\cdot 3}+....+\dfrac{11-10}{11\cdot 10}=[/tex]

[tex]\bf \dfrac{2}{2\cdot 1}-\dfrac{1}{2\cdot 1}+\dfrac{3}{3\cdot 2}-\dfrac{2}{3\cdot 2}+\dfrac{4}{4\cdot 3}-\dfrac{3}{4\cdot 3}+....+\dfrac{11}{11\cdot 10}-\dfrac{10}{11\cdot 10}=[/tex]

[tex]\bf 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-....-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}=[/tex]

[tex]\bf 1-\not\dfrac{1}{2}+\not\dfrac{1}{2}-\not\dfrac{1}{3}+\not\dfrac{1}{3}-\not\dfrac{1}{4}+\not\dfrac{1}{4}-....-\not\dfrac{1}{10}+\not\dfrac{1}{10}-\dfrac{1}{11}=[/tex]

[tex]\bf 1-\dfrac{1}{11}=[/tex]

[tex]\bf \dfrac{11-1}{11}=[/tex]

[tex]\bf \dfrac{10}{11}[/tex]