Question

1.Fie s=1+4+4²+.....+4^{2015
Calculati 3s+1.
2.Determinati valorile nat ale nr n si cifra x pentru care are loc egalitatea:
3n+6+3n+5+3 n+4 +2*3 n+3 4*3 n=xxxx(cu bara de asupra)

Answer 1

S=1+4+$4^{2} +....+ 4^{2015}$ I*4
4S= 4+$4^{2} + 4^{3} +....+ 4^{2016}$
se scad cele 2 rel
3S=[tex] 4^{2016} -1 [/tex]
3S+1=$4^{2016}$
2. Nu stiu daca ai scris bine ex. dar pt ce ai scris rezolvarea este:
3n+6+3n+5+3n+4+6n+102n=1000x+100x+10x+x
117n+15=1111x
x cifra,x≠0
pt x=1⇒117n+15=1111⇒117n=1111-15⇒117n=1096⇒n=1096/117∉N
pt x=2⇒117n+15=2222⇒117n=2222-15⇒n=2207/117∉N
si tot asa pana la 9.