how to solve tis one: mathematical induction (1 Viewer)

[lie_stella]

5^n > 3^n +2^n thanks!

 

[shafqat]

5^n > 3^n +2^n thanks!

5^n = (3+2)^n
= 3^n + 3^n-1.2 + ..... + 2^n
> 3^n + 2^n, for n>1

 

[Ghost1788]

hey could you explain the 2nd line..someone... thanks

 

[lucifel]

i think he just forgot the nCr's, but otherwise it follows.

 

[shafqat]

^ = to the power of
Here's the induction proof:
5^n > 3^n +2^n
Assume 5^k > 3^k +2^k
For k+1, LHS = 5^k+1
= 5.5^k
> 5(3^k + 2^k)
= 5.3^k + 5.2^k
> 3.3^k + 2.2^k
= RHS

The other proof is just through binomial expansion.

 

[gman03]

5ⁿ⁺¹
= 5 . 5ⁿ
> 5 . ( 3ⁿ +2ⁿ) by induction
= 5 . 3ⁿ + 5 . 2ⁿ
> 3 . 3ⁿ + 2 . 2ⁿ
= 3ⁿ⁺¹ +2ⁿ⁺¹


for n = 2, 3² + 2² = 13 < 25 = 5²

Hence, by induction, 5ⁿ > 3ⁿ +2ⁿ is true for n >= 2.

EDIT: A bit too late, it seems

 

[Ghost1788]

Cant you just say

5^n = (3 + 2)^n

=3^n + 2^n + 12
therefore

5^n > 3^n + 2^n

Similarly

5^(n+1) = (3 + 2)^(n+1)
=3^(n+1) + 2^(n+1) + 12
therefore

5^(n+1) = 3^(n+1) + 2^(n+1)

By Mathematical Induction

plz correct me if i'm wrong

 

[shafqat]

Cant you just say

5^n = (3 + 2)^n

=3^n + 2^n + 12

That step is incorrect.

 

[Ghost1788]

yea i just realised this is only applicable for n=2...hmm i am a lost cause i think...