" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a)
an= (2n+1)²- [2(n-1)+1]²= 4n²+4n+1-[2n- 1]²= 4n²+4n+1-[4n²- 4n+ 1]=
4n²+ 4n+ 1- 4n²+ 4n- 1= 8n
an= 8n
b)
an= [n (n+1 )]/ 2- { [ (n-1)n ]/2 }= [n²+ n]/2- { [ n²- n ]/2 }=
= [n²+ n- n²+ n]/2= 2n/ 2= n
an= n
c)
an= (n+ 1 )/ (n+ 2 ) - [ (n )/ (n+ 1 ) ]= [(n+ 1)²- n(n+ 2)]/ [(n+ 2)(n+1)]= [n²+ 2n+ 1- n²- 2n]/ [(n+ 2)(n+ 1)]= 1/[(n+ 2)(n+ 1)]
an= 1/[(n+ 2)(n+ 1)]
//mam nadzieje ze dobrze