Answer:
a) 4(9
+2)
b)6(7
+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;
= ![3^{2n} -1](https://tex.z-dn.net/?f=3%5E%7B2n%7D%20-1)
= ![3^{2\times1} -1](https://tex.z-dn.net/?f=3%5E%7B2%5Ctimes1%7D%20-1)
= 9 - 1
= 8
hence it is divisible by 4
Let the statement is for n = k
∴
= 4
(equation 1)
= 4![x](https://tex.z-dn.net/?f=x)
= 4
(equation 1)
Now, we have to proof the statement is true for n = k+1
= ![3^{2(k+1)} -1](https://tex.z-dn.net/?f=3%5E%7B2%28k%2B1%29%7D%20-1)
=
(
)
Adding & Subtracting 8
= ![(3^{2k} \times 3^{2} ) -1 +8 -8](https://tex.z-dn.net/?f=%283%5E%7B2k%7D%20%5Ctimes%203%5E%7B2%7D%20%29%20-1%20%2B8%20-8)
= ![9^{k} \times 9 -9 + 8](https://tex.z-dn.net/?f=9%5E%7Bk%7D%20%5Ctimes%209%20-9%20%2B%208)
taking common 9
= 9(
-1)+8
= 9 (4
) +8 (from equation 1)
= 36
+ 8
= 4(9
+2)
if (9
+2) = p
then = 4p
Since
= 4p evenly divisible by 4
therefore given statement is true
b)Prove that for any positive integer n, 6 evenly divides ![7^{n} - 1](https://tex.z-dn.net/?f=7%5E%7Bn%7D%20-%201)
checking whether the statement is correct or not
∴ n = 1;
![7^{n} - 1](https://tex.z-dn.net/?f=7%5E%7Bn%7D%20-%201)
7 - 1
6
6 is divisible by 6
hence the given statement is true for n = 1
let it also true for n = k
(equation 2)
Now we have to proof the statement is true for n = k+1
![7^{k+1} - 1](https://tex.z-dn.net/?f=7%5E%7Bk%2B1%7D%20-%201)
![7^{k}\times7 - 1](https://tex.z-dn.net/?f=7%5E%7Bk%7D%5Ctimes7%20%20-%201)
Adding & Subtracting 6
![7^{k}\times7 - 1 +6 - 6](https://tex.z-dn.net/?f=7%5E%7Bk%7D%5Ctimes7%20%20-%201%20%2B6%20-%206)
![7^{k}\times7 - 7 +6](https://tex.z-dn.net/?f=7%5E%7Bk%7D%5Ctimes7%20%20-%207%20%2B6)
![7(7^{k}\times - 1) +6](https://tex.z-dn.net/?f=7%287%5E%7Bk%7D%5Ctimes%20%20-%201%29%20%2B6)
7(6
)+6 ( from equation 2)
= 42
+ 6
= 6(7
+1)
if 6(7
+1) = p
then = 6p
Since
= 6p evenly divisible by 6
therefore given statement is true