Răspuns:
a)a = 2017 + 2(1 + 2 + ... + 2016)a)a=2017+2(1+2+...+2016)
a = 2017 + 2 \times \frac{2016(2016 + 1)}{2} a=2017+2×
2
2016(2016+1)
a = 2017 + 2016(2016 + 1)a=2017+2016(2016+1)
a = 2017 + 2016 \times 2017a=2017+2016×2017
a = 2017(1 + 2016)a=2017(1+2016)
a = 2017 \times 2017a=2017×2017
a = {2017}^{2} = > p.pa=2017
2
=>p.p
b)b = 1 + 3 + 5 + ... + 2017b)b=1+3+5+...+2017
1 + 3 + 5 + ... + 2n - 1 = {n}^{2} 1+3+5+...+2n−1=n
2
2n - 1 = 20172n−1=2017
2n = 2017 + 12n=2017+1
2n = 2018 \: | \div 22n=2018∣÷2
n = 1009n=1009
{n}^{2} = {1009}^{2} = > p.pn
2
=1009
2
=>p.p
c)c = 81 + 2 \times 81 + 3 \times 81 + ... + 49 \times 81c)c=81+2×81+3×81+...+49×81
c = 81(1 + 2 + 3 + ... + 49)c=81(1+2+3+...+49)
c = 81 \times \frac{49(49 + 1)}{2} c=81×
2
49(49+1)
c = 81 \times \frac{49 \times 50}{2} c=81×
2
49×50
c = 81 \times 49 \times 25c=81×49×25
c = {9}^{2} \times {7}^{2} \times {5}^{2} c=9
2
×7
2
×5
2
c = {(9 \times 7 \times 5)}^{2} = > p.pc=(9×7×5)
2
=>p.p
d)d = 2(1 + 2 + 3 + ... + 124) + 125d)d=2(1+2+3+...+124)+125
d = 2 \times \frac{124(124 + 1)}{2} + 125d=2×
2
124(124+1)
+125
d = 124(124 + 1) + 125d=124(124+1)+125
d = 124 \times 125 + 125d=124×125+125
d = 125(124 + 1)d=125(124+1)
d = 125 \times 125d=125×125
d = {5}^{3} \times {5}^{3} d=5
3
×5
3
d = {(5 \times 5)}^{3} d=(5×5)
3
d = {25}^{3} = > c.pd=25
3
=>c.p
Sper ca te-am ajutat!
