F.4 AM

(1)



Let T(n) = [(3n+1)7^n] - 1 where n is all natural number



For n = 1, T(1) = {[3(1)+1]7^(1)} -1=27



T(1) is divisible by 9



Assume T(k) is divisible by 9 where k is a natural number



T(k) = [(3k+1)7^k] - 1



if n = k+1,



T(k+1) = {[3(k+1)+1]7^(k+1)} - 1



= [7(3k+4)7^k] - 1



= [7(3k+1)7^k +21(7^k)] - 1



= 7[(3k+1)7^k-1] + 21(7^k)+6



= 7[T(k)] + 21[(3k+1)7^k -1] -21(3k)(7^k) +27



= 7[T(k) + 21[T(k)] -63k(7^k) + 27



= 28T(k) - 63k(7^k) + 27



T(k) , 63k(7^k) and 27 are divisible by 9



Therefore , T(k+1) is divisible by 9.



Therefore, by M.I., T(n)= (3n+1)7^n - 1 is divisible by 9 where n is all natural number.







Let T(n) = 7^2n - 48n - 1 where n is all natural number



For n = 1, T(1) = 7^2(1) - 48(1) - 1 = 0



T(1) is divisible by 2304



For n = 2,



T(2) = 7^2(2) - 48(2) - 1



= 2401-96-1=2304



T(2) is divisible by 2304



Assume T(k) is divisible by 2304 where k is a natural number



T(k) = 7^2k - 48k - 1



If n = k+1



T(k+1) = 7^2(k+1) - 48(k+1) - 1



= 49(7^2k) - 48k - 49



= 49(7^2k - 48k - 1) +2304 k



= 49[T(k)] + 2304k



T(k) and 2304k are divisible by 2304.



Therefore, T(k+1) is divisible by 2304.



By M.I. , T(n) = 7^2n - 48n - 1 is divsible by 2304 where n is all natural nnumber

1) let s(n) be the statement"(3n+1)7^n-1 is divisble by 9",
when n = 1
LHS,4*7 - 1 = 27 which is divisible by 9
S(1) IS TRUE
Assume s(k) is true,
(3k+1)7^k -1 = 9m where m is an integer.
when n = k+1,
LHS = (3(k+1)+1)7^(k+1) - 1
= (3k+4)7*7^k -1
= (21k + 28)7^k -1
= (21k+28)7^k + 9m - (3k+1)7^k
= (21k-3k + 28 -1)7^k + 9m
= 126k^2 +189k + 9m
= 9(14k^2 + 21k + m)
s(k+1) is true
by the principle of mi,(3n+1)7^n-1 is divisble by 9 for all natural numbers n.


2)let s(n) be the statement"7^2n-48n-1 is divisble by 2304 ",
when n = 1
LHS,7^2 - 48 - 1 = 0 which is divisible by 2304
S(1) IS TRUE
Assume s(k) is true,
7^2k-48k-1 = 2304m where m is an integer.
when n = k+1,
LHS =7^2(k+1) - 48(k+1) -1
=7^2k * 7^2 -48k -48 -1
= 49(7^2k) +2304m - 7^2k - 48
= 48(2304m + 48k - 1) + 2304m - 48
= 48(2304m + 48k - 1 + 48m-1)
= 48(2352m + 48k)
= 48(48(49m+k))
= 2304(49m+k)
s(k+1) is true
by the principle of mi,7^2n-48n-1 is divisble by 2304 for all natural numbers n.