1(an + bn ) > [(a+b) n]/2
I think we should us this inequality.
which is greater?
99^n + 100^n or 101^n???
If you get ambiguous answers then specify the intervals for which 99^n+100^n is greater than 101^n or less than it.
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8 Answers
341hint: try to prove that 101n -99n>100n or
(1 + 1/100)n - (1-1/100)n> 1
106see 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
341It 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
106could u please show the steps. i mean how did u determine that the sign changes at n=49
341sorry, 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
106can u use the same logic and find out whether 101^48>100^48+99^48