Question

a) Verificați următoarele egalități: b) calculați următoarele sume : ​

Answer 1

a) Verificati urm. egalitati:

$i) \frac{1}{k} -\frac{1}{k+1} =\frac{k+1}{k(k+1)}- \frac{k}{(k+1)k} =\frac{k+1-k}{k(k+1)} =\frac{1}{k(k+1)} \\ii) \frac{1}{2} [ \frac{1}{k} -\frac{1}{k+2} ]=\frac{1}{2}[\frac{k+2}{k(k+2)} -\frac{k}{(k+2)k} ] =\frac{1}{2} * \frac{k+2-k}{k(k+2)} =\frac{1}{2} * \frac{2}{k(k+2)} =\frac{1}{k(k+2)} \\iii)\frac{1}{2} [\frac{1}{k(k+1)} -\frac{1}{(k+1)(k+2)} ]= \frac{1}{2} [\frac{k+2}{k(k+1)(k+2)} -\frac{k}{k(k+1)(k+2)} ]=\frac{1}{2} * \frac{2}{k(k+1)(k+2)}=\frac{1}{k(k+1)(k+2)}$

b) Calculati urm. sume:

$i) S1= \frac{2-1}{1*2} + \frac{3-2}{2*3} +\frac{4-3}{3*4} + ...+\frac{100-99}{99*100} \\S1=\frac{2}{1*2} -\frac{1}{1*2} +\frac{3}{2*3} -\frac{2}{2*3} +\frac{4}{3*4}-\frac{3}{3*4} +...+\frac{100}{99*100} -\frac{99}{99*100} \\S1=1-\frac{1}{2} +\frac{1}{2} -\frac{1}{3} +\frac{1}{3} -\frac{1}{4} +...+\frac{1}{99} -\frac{1}{100} \\S1=1-\frac{1}{100} =\frac{100-1}{100} =\frac{99}{100}$$ii) \frac{1}{1*3} =\frac{1}{2}[\frac{1}{1} -\frac{1}{3} ];\\\frac{1}{2*4} =\frac{1}{2} [\frac{1}{2} -\frac{1}{4} ]; \\deci\\S2=\frac{1}{2} [\frac{1}{1} -\frac{1}{3} +\frac{1}{2} -\frac{1}{4} +\frac{1}{3} -\frac{1}{5} +\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98} -\frac{1}{100} ]\\S2=\frac{1}{2} *[1-\frac{1}{100} ]=\frac{1}{2} * S1$

Folosind acelasi procedeu ca la subpunctul ii), $iii) S3=\frac{1}{2} [\frac{1}{1*2} -\frac{1}{2*3}+\frac{1}{2*3} -\frac{1}{3*4} +...+\frac{1}{98*99} -\frac{1}{99*100} ]=\frac{1}{2} *[\frac{1}{2}-\frac{1}{99*100} ]$