Three induction questions (1 Viewer)

[sinophile]

 

[lyounamu]

x^n - 1 by (x-1)

when n=1,

x^n -1 = x -1 = (x-1) x 1 therefore it is divisible by (x-1) when n=1

assume that it is also true for n=k

i.e. x^k -1 = (x-1)k where k is another integer

now prove that it is true for n=k+1

i.e. x^(k+1) - 1 = T(x-1) where T is another integer

LHS = x^k . x - 1 = x^k . x - x + x -1 = x(x^k -1 ) -1
= x . k(x-1) - 1 + x
= (x-1) (kx +1) = (x-1)T where T = kx+1
= RHS

proved true by the principle of mathematical induction

 

[gurmies]

 

[jet]

I'll do 7) When n = 2, LHS = 144, RHS = 49 + 25 = 74
Therefore true for n = 2
Now, assume true for n = k, i.e assume 12^k > 7^k + 5^k
We now prove for n = k+ 1 i.e that 12^{k+ 1} > 7^k+1 + 5^k+1
LHS = 12 x 12^k
> 12 (7^k + 5^k)
= 12 x 7^k + 12 x 5^k
> 7 x 7^k + 5 x 5^k
= 7 ^k+1 + 5^k+1
= RHS
Hence, if it is true for n=k, then it is true for n = k+1. Since it is true for n = 2, then it is true for n=3,4,5, hence it is true for n≥2

 

[lyounamu]

6.

7^n + 19^n

n = 1

7 + 19 = 26 and this is divisibly by 13

now assume n=k for k=odd pos number

7^k + 19^k = 13G where G is another integer

prove that n=k+2

7^(k+2) + 19^(k+2) = 13T where T is another integer

LHS = 49 . 7^k + 361 . 19^k = 49 . 7^k + 49 . 19^k + 312 . 19^k
= 49 (13G) + 312 . 19^k
= 13 (49G + 24 . 19^k)
= 13T
= RHS

proved true by the mathematical induction

EDIT: I also did n=1, and it was proven wrong meaning that it cannot be even.. (cbf putting that one)