Respuesta :
Answer:
a) 4(9[tex]x[/tex]+2)
b)6(7[tex]x[/tex] +1)
Step-by-step explanation:
a) Prove that for any positive integer n, 4 evenly divides 32n-1
checking whether the statement is correct or not
∴  n = 1;
= [tex]3^{2n} -1[/tex]
= [tex]3^{2\times1} -1[/tex]
= 9 - 1
= 8
hence it is divisible by 4
Let the statement is for n = k
∴  [tex]3^{2k } -1[/tex]  = 4[tex]x[/tex](equation 1)
[tex](3^{2})^{k} -1[/tex] = 4[tex]x[/tex]
[tex]9^{k} -1[/tex] = 4[tex]x[/tex] (equation 1)
Now, we have to proof the statement is true for n = k+1
= [tex]3^{2(k+1)} -1[/tex]
= [tex](3^{2k} \times 3^{2} ) -1[/tex] ([tex]x^{a+b} = x^{a} \times x^{b}[/tex])
Adding  & Subtracting 8
= [tex](3^{2k} \times 3^{2} ) -1 +8 -8[/tex]
= [tex]9^{k} \times 9 -9 + 8[/tex]
taking common 9
= 9([tex]9^{k}[/tex] -1)+8
= 9 (4[tex]x[/tex]) +8 (from equation 1)
= 36[tex]x[/tex] + 8
= 4(9[tex]x[/tex]+2)
if (9[tex]x[/tex]+2) = Â p
then = 4p
Since [tex]3^{2(k+1)} -1[/tex] = 4p evenly divisible by  4
therefore given statement is true
b)Prove that for any positive integer n, 6 evenly divides [tex]7^{n} - 1[/tex]
checking whether the statement is correct or not
∴  n = 1;
[tex]7^{n} - 1[/tex]
7 - 1
6
6 is divisible by  6
hence the given  statement is true for n = 1
let it  also true  for n = k
[tex]7^{k} - 1 = 6x[/tex] (equation 2)
Now we have to proof the statement is true for n = k+1
[tex]7^{k+1} - 1[/tex]
[tex]7^{k}\times7 - 1[/tex]
Adding  & Subtracting 6
[tex]7^{k}\times7 - 1 +6 - 6[/tex]
[tex]7^{k}\times7 - 7 +6[/tex]
[tex]7(7^{k}\times - 1) +6[/tex]
7(6[tex]x[/tex] )+6 ( from equation 2)
= 42[tex]x[/tex] + 6
= 6(7[tex]x[/tex] +1)
if 6(7[tex]x[/tex] +1) = Â p
then = 6p
Since[tex]7^{k+1} -1[/tex] = 6p evenly divisible by  6
therefore given statement is true
