Question

subpunctul b) S2..........

Answer 1

1/(√2+2√1)=(√2-2√1)/(√2+2√1)(√2-2√1)=(√2-2√1)/(2-4)=(√2-2√1)/(-2)=
=-√2/2+√1/1
1/(2√3+3√2)=(2√3-3√2)/(2√3+3√2)(2√3-3√2)=(2√3-3√2)/(12-18)=
=(2√3-3√2)/(-6)=-√3/3+√2/2
1/(3√4+4√3)=(3√4-4√3)/(3√4+4√3)(3√4-4√3)=(3√4-4√3)/(36-48)=
=(3√4-4√3)/(-12)=-√4/4+√3/3


1/[n(√(n+1)+(n+1)√n)]=[n(√(n+1)-(n+1)√n]/[n²(n+1)-(n+1)²×n]=
=[n(√(n+1)-(n+1)√n]/[n(n+1)(n-n-1)]=[n(√(n+1)-(n+1)√n]/[-n(n+1)]=
=-[√(n+1)]/(n+1)+√n/n

1/(√2+2√1)+1/(2√3+3√2)+....+1/[n(√(n+1)+(n+1)√n]=
=-√2/2+√1/1-√3/3+√2/2-√4/4+√3/3........-[√(n+1)]/(n+1)+√n/n=
=√1/1+(√n)/n=1+(√n)/n=(n+√n)/n