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By the principal of mathematical induction prove that n^2>2n where n is greater than or equals to 3.

one year ago

Arun
23368 Points

Suppose n=3

then 3^2=9>2*3=6

Now let n=4

4^2=16>8

Now let’s look at n+1

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

n^2+2n+1>2(n+1)

n^2>1 which is true for every n greater than 2 or 2. As we looked at n+1: 2+1=3 thus it is true for n>=3.

QED

Also if n is real, not natural there is another way to proove this

Let’s look at the derivatives of f(n)=n^2 and g(n)=2n (functions are continuous R->R)

f’(n)=2n; g’(n)=2

Values of them are equal only in one point if n=1.

If n>1 than monotonicaly increasing 2n is bigger than constant function 2.

The original functions intersect in

n^2=2n

n=0; n=2

Functions f and g have the same values in n=2 but f’ is bigger. That means they start from one point, but n^2 always increases faster for n>2. Thus, it will have bigger values and the gap will be increasing. Lim(n^2/2n) n->inf diverges.

one year ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions