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2 Let n be positive integer; and let < 6 < 6 < c < n. Prove that (a,6)(b,c) (a,6) (b,c) (a,6)(b,c)_...

Question

2 Let n be positive integer; and let < 6 < 6 < c < n. Prove that (a,6)(b,c) (a,6) (b,c) (a,6)(b,c)_

2 Let n be positive integer; and let < 6 < 6 < c < n. Prove that (a,6)(b,c) (a,6) (b,c) (a,6)(b,c)_



Answers

Prove that the statement is true for every positive integer $n$.
If $0<a<b,$ then $\left(\frac{a}{b}\right)^{n+1}<\left(\frac{a}{b}\right)^{n}$.

First Equals one. This statement is A over B squared. It's Mourners and hey over B. Since A. And be a positive A or B is also politics and they is equivalent to A or b. s. monitor than one since A is smaller than be. This is true. Next we assume that this statement is true for any calls cape. Then for ecos K plus one it becomes A over B. The power of okay plus two. He calls a over B to the power of Okay plus one times. Hey over B which is smaller than yeah, A over B to the power of cake times A or B. Note that this is because the statement is true for ecowas cake and enhance A over B to the power of Okay plus one is smaller than A or B to the power of case. And has this equals A. Or B to the power of K plus one. Which means the statement is true for any equals K plus one. Yes. By mathematical inducting. The statement is true for every and which positive industry

In this problem of basics of contemporary we have to prove each statement for positive and teasers are and and where are less than are required to end. And we have given combination In Coma in -1 equal to and taking alleges We help combination and comma and -1 and we have from love food combination C N comma are equal to and pictorial upon and minus R. Factorial. Multiplying with our victoria lee. Using Formula We Compute CNO in -1. So we have N factorial upon in minors in minus one sectorial multiplying. Wait 8 -1 Factorial. First we simplify the expression so we have N factorial upon one factorial Multiplying in -1 factorial. Using formula for N. Factorial And multiplying and -1 factorial. We expand this in victoria and we know value of one factorial is equal to one. So we have And in -1 factorial upon and minors one factorial. Now we can divide this in -1 factorial With this end -1 factorial. So we have and here alleges equal to end and RHS is also equal to mm. So we can say here, L H is equal to our ages has fruit. This is our final answer.

In this problem of basics of counting theory we have to prove each statement for positive and desert and and are. We're in less than or equal to end and we have given combination and comma and miners are equal to combination and comma are first we take alleges and use the formula for combination and coma. It's weird too. And factorial upon in minus R. Factorial multiplying with R. Factorial. Now we put value of LHs into the farm lock so we have combination and comma and minus are equal to and pictorial upon and minus and minus R. Whole factorial multiplying wait and minus R. Factorial. Simplifying the expression we have value and factorial upon R. Factorial. Multiplying weight in minus R. Factorial. Now we take our riches and use the combination formula to find the value. So I'm putting the value we have and factorial upon and minus R. Factorial. Multiplying with our pictorial these LHs Question one acknowledges equation too. So we can say from a question one and from a question too, LHs is equal to Rhs as through this is our final answer

Hello. So here are given sequence is the end fruit of eight to the end plus B to the end. So if you take the limit as N goes to infinity of the end. Fruit of a de emphasis to the end. While we have an indeterminate form which is going to be infinity to the zero. Can then apply low petals role. Um And we take again the limit of well this is going to give us the limit of um age of the fsb to the end to the one over. And which is going to give us the limit as N approaches infinity of we get here um A A to the end times Ln of A plus B to the end times Ln of be all over A. To the N. B to the end. Which is going to give us here A A over B to the N. Times Ln of A plus Ln of B. All divided by A over B. To the N. Um plus one. And then um taking the limit here, this is gonna gonna just convert to us to the natural log of BE. So therefore yes, we take the limit um As N. Goes to infinity of the natural log of eight to the end to be to the end to the one over end is going to be equal um and converged to be.


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