Question

Determinați numărul natural n din egalitatile :
a)7^20+24+28+.....+48=49n
b)13^1+3+5+7+......+49=13^n^2
c)5^n+3+5^n+1-15^n:3^n=29×25^13
d)9^n+1×9^2n+2=27^14
urgent

Answer 1

a)7^(20+24+28+.....+48)=49^n
nr termeni (48-20):4+1=28:4+1=7+1=8
20+24+28+.....+48=20+(20+4)+(20+8)+...+(20+28)=20×8+(4+8+...+28)=
=160+4(1+2+...+7)=160+4×7×8/2=160+112=272
7^272=7^2n
2n=272
n=136

b)13^(1+3+5+7+......+49)=13^n^2
1+3+5+7+......+49=1+2+3+4+5+6+7+....+48+49-2-4-6-....-48=
=49×50/2-2(1+2+3+....+24)=49×25-2×24×25/2=1225-600=625
13^625=13^n^2
n^2=625
n=√625=25

c)aici am facut 2 variante  (in loc de 29 cred ca trebuia 129   sau primul 5 e la puterea n+2)
5^n+3+5^n+1-15^n:3^n=129×25^13
5^n+3+5^n+1-5^n=129×5^26
5^n(5^3+5^1-5^0=129×5^26
5^n(125+5-1)=129×5^26
5^n ×129=129×5^26
n=26

5^n+2+5^n+1-15^n:3^n=29×25^13
5^n+2+5^n+1-5^n=29×5^26
5^n(5^2+5^1-5^0=29×5^26
5^n(25+5-1)=29×5^26
5^n ×29=29×5^26
n=26

d) 9^n+1×9^2n+2=27^14
3^2n+2×3^4n+4=3^42
3^6n+6=3^42
6n+6=42
6n=36
n=6