Question

Determinati numarul natural n din egalitate :
a)2 la 1 +2+3...+16+4 la n+7
b)4 la a 2 +4+6+....+20=4 la n (n+1 )
c)5 la a 11+12+13+..30+(5 la n) la a treila ori 25
d)7 la 20 +24+28...+48=49 la n
e )2 la 5 +12 +19+...+75+16 la 11n
f)13 la 1 +3+5 +7 +...+49=13 la n la a 2

Answer 1

a)2¹⁺²⁺³⁺⁴⁺     ⁺¹⁶=(2²)ⁿ⁺⁷⇒2¹³⁶=2²⁽ⁿ⁺⁷⁾  136=2(n+7)⇒68=n+7⇒n=61
apl. suma gauss  1+2+3+...+16=16*17/2=136

b)(2²)²⁺⁴⁺⁶⁺         ⁺²⁰=(2²)ⁿ⁽ⁿ⁺¹⁾⇒2²²⁰=2²ⁿ⁽ⁿ⁺¹⁾⇒220=2n(n+1)⇒110=n(n+1)
(110=n²+n       m am incuiat,poate ai tu vreo idee)
apl. SG   2(1+2+3+...+10)=2x10x11/2=110

c) SG   (1+2+3+................+30)-(1+2+3+.........+10)=30*31/2-10*11/2=465-55=410
(5ⁿ)³x5²
5⁴¹⁰=(5ⁿ)³x5²
410=n³+2       n=∛408 dc.faci verificarea e ok!

d)SG    4(5+6+7+....+12)=4[(1+2+.......12)-(1+2+3+4)]=4[78-10]=4*68=272
7²⁷²=7²ⁿ        272=2n    n=136

e) aflam nr.de termeni (75-5):7+1=11termeni
nu mi amintesc formula SG ptr. aceasta suma,mai caut si dc.gasesc termin ex.

f) SG   (1+2+3+4+..........+49)-2(1+2+3+...........+24)=49*50/2-2*24*25/2=1225-600=625
13⁶²⁵=(13ⁿ)²⇒625=n²⇒n=25

Answer 2

rezolvarea in atasament