Question

Salut! Proiect de recuperare temă de algebră: exerciţiul 24, pct. a şi b

Answer 1

Rezolvările sunt în atașamente.

Numărătorul se scrie ca o diferență dintre factorii numitorului.

Pentru ambele cazuri termenii se reduc, rămânând primul și ultimul.

Answer 2

$24) \\ \\ a)~~\dfrac{3}{1\cdot 4}+\dfrac{3}{4\cdot 7}+\dfrac{3}{7\cdot 10}+...+\dfrac{3}{97\cdot 100} = \\ \\ =\displaystyle \sum\limits_{k=1}^{33}\dfrac{3}{(3k-2)(3k+1)}= \sum\limits_{k=1}^{33}\dfrac{(3k+1)-(3k-2)}{(3k-2)(3k+1)} = \\ \\ =\sum\limits_{k=1}^{33}\Big(\dfrac{1}{3k-2}-\dfrac{1}{3k+1}\Big) = \\ \\ = \dfrac{1}{1}+\dfrac{1}{4}+\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{4}-\dfrac{1}{7}-...-\dfrac{1}{97}-\dfrac{1}{100} = \\ \\ = 1-\dfrac{1}{100} = \dfrac{99}{100}$

$b)~~\dfrac{11}{1\cdot 12}+\dfrac{11}{12\cdot 23}+\dfrac{11}{23\cdot 24}+...+\dfrac{11}{89\cdot 100} = \\ \\ =\displaystyle \sum\limits_{k=1}^{9}\dfrac{11}{(11k-10)(11k+1)}= \sum\limits_{k=1}^{9}\dfrac{(11k+1)-(11k-10)}{(11k-10)(11k+1)} = \\ \\ =\sum\limits_{k=1}^{9}\Big(\dfrac{1}{11k-10}-\dfrac{1}{11k+1}\Big) = \\ \\ = \dfrac{1}{1}+\dfrac{1}{12}+\dfrac{1}{23}+...+\dfrac{1}{89}-\dfrac{1}{12}-\dfrac{1}{23}-...-\dfrac{1}{89}-\dfrac{1}{100} = \\ \\ = 1-\dfrac{1}{100} = \dfrac{99}{100}$