Question

luind in consideratie ca 1/n - 1/n+1 = 1/n(n + 1) pentru orice n apartine nr naturale nenule, caculati 1/1x2 + 1/2x3 + ... + 1/99x100

Answer 1

$\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\ldots+\dfrac{1}{99\cdot 100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}=\\\\=1-\dfrac{1}{100}=\dfrac{99}{100}.$

Green eyes.