inequality

(aⁿ + bⁿ ) > [(a+b) ⁿ]/2

I think we should us this inequality.

hint: try to prove that 101ⁿ -99ⁿ>100ⁿ or

(1 + 1/100)ⁿ - (1-1/100)ⁿ> 1

see i think i know at u mean 2 say but see for n=1,2
101^n<100^n + 99^n

but check for n = 50 u get 101^n > 100^n + 99^n

I want to know for which value of n the inequality sign changes and the process also

It changes sign at n = 49. Thereafter the inequality holds as it is an increasing function

Actually, the version i had seen earlier was to prove this for n≥49

sorry, i didnt mean the sign changes at n=49. Like I said, we can prove that it is positive for n = 49 and since it is an increasing function, we can prove that for n≥49 it is +ve. Thats how it was in a question i solved earlier, and most likely that's the best presentation of the problem.

(1+1/100)⁴⁹ - (1-1/100)⁴⁹ = 2(⁴⁹C₁ 1/100 + p]49C₃ 1/100³ +...(+ve terms)

> 2 X 0.49 + 2 ⁴⁹C₃ 1/100³

So now we have to prove that 2 ⁴⁹C₃ 1/100³>0.02

After simplifying, we have to prove that 49X47>1250 which is easily verified

So the inequality hold for n = 49.

Now to the part of proving that the sequence is an increasing one.

(1+1/100)ⁿ increases with n

(1-1/100)ⁿ decreases with n. But -(1-1/100)ⁿ increases with n.

So (1+1/100)ⁿ-(1-1/100)ⁿ is therefore increasing and hence >1 for n≥49