2 inequality qn (1 Viewer)

[blakwidow]

Show by maths induction that

3ⁿ > 1 + 2n
.....-

(1+p)ⁿ > 1+ np, where p > -1
............-

n in both cases is true for all values?
What do i start off the statement with? n=1?

 

[vafa]

p(n): 3^n>1+2n
p(k-1): 3^(k-1)>2k-1
p(k): 3^k>1+2k
3.p(k-1), 3^k>6k-3
6k-3>1+2k
hence 3^k>1+2k


p(n): (1+p)^n>1+np
p(k-1): (1+p)^(k-1)>1+pk-p
p(k): (1+p)^k>1+kp
(1+p).p(k-1), (1+p)^k>1+pk+p^2(k-1)
1+pk+p^2(k-1)>1+kp
hence (1+p)^k>1+kp

in both question you need to start off with n>1
you could start with n>=1, if you also have the inequality in both cases.

 

[pLuvia]

For both cases it would only be true for n>1

Then do all the normal algebra bashing

 

[blakwidow]

p(n): 3^n>1+2n
p(k-1): 3^(k-1)>2k-1
p(k): 3^k>1+2k

in both question you need to start off with n>1
you could start with n>=1, if you also have the inequality in both cases.

If you say P(k) : ...
P(k+1) ...

does that include the negative numbers too?

 

[vafa]

I really did not understand your point.
but you are the one who decides choosing k or k-1 for assumtion.
if you start with k, then need to prove for k+1
k-1 does not mean that we can include negative numbers.

 

[blakwidow]

How do u prove its true for negative integers too?

 

[pLuvia]

Depends if the question wants you to prove it for all n or a restricted domain for n

 

[vafa]

I am sure the principle of mathematical induction only works for natural numbers (usually).

To clarify it please look at this document provided by the Department of Mathematics of Oxford University.
please pay special attention to pages 3 and 4.

http://www.maths.ox.ac.uk/prospective-students/undergraduate/single-a-level/induction/induction.pdf

 

[blakwidow]

ok thanks for the help