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Fie fct f(x) = 2x -1
calculati f(1)+f(2)+...+f(2013)


Răspuns :

f(1)=2*1-1
f(2)=2*2-1
f(3)=2*3-1
...
f(n)=2*n-1
f(1)+f(2)+...+f(n)=2*1-1+2*2-1+...+2*n-1
=2*1+2*2+2*3+...2*n-(1+1+...+1) (1 este de n ori)
=2(1+2+...+n)-n=2*n(n+1)/2 -n =n(n+1)-n=n(n+1-1)=n^2
*-operatia de inmultire
^-ridicarea la putere
1+3+5+...+4023+ 4025=4026*2013/2=2013*2013=2013²=4052169

ptca am 2013 numere deci 2013/2 perechi cu suma 1+4025=4026


ALTFEL
2*1-1
2*2-1
...........
2*2013-1

2(1+2+....+2013)-1-1...-1 (2013de 1)=
2*2013*2014/2-2013=
2013*2014-2013=
2013(2014-1)=
2013*2013=2013²=4052169