Prove using induction that for all integers n >5, 4n<2^n?
There are three parts to the inductive proof: I) prove true for the smallest value of n: 4(6)<2^6 24<64 True for n=6 II). Assume true for some value of n=k 4(k)<2^k Iii) prove true for n= k+1: 4(k+1)<2^(k+1) K>5 4k>20>4 4k<2^k 2*4k < 2*2^k 2*4k< 2^(k+1) 4k+ 4k < 2^(k+1) 4k+ 4< 4k+4k < 2^(k+1) 4(k+1)< 2^(k+1) Hoping this helps!
Use induction to prove that for each natural number n, 1^3+2^3+3^3+...n^3=(n(4n^2-1))...
The case n=1 is straightforward: 1^3=1=(1*(1+1)/2) We now suppose we proved the formula for n, and we prove it for n+1: 1^3+2^3+...+n^3+(n+1)^3 = (n(n+1)/2)^2+(n+1)^3 = (n^4+3n^2+3n)/4+n^3+3n^2+3n+1 = (n^4 + 2n^3 + n^2 + 4n^3 + 12n^2 + 12n + 4)/4 = (n^4 + 6n^3 + 13n^2 + 12n + 4)/4 = ((n+1)(n+2)/2)^2
Use induction to prove that for each natural number n, 1^2+2^2+3^2+...n^2=(n(n+1)(2n+...
Using induction prove for the natural numbers n 1^2+2^2+3^2+...n^2=(n(n+1)(2n+1))/6 pf if n=1 then clearly 1^2 = (1)(2)(3)/6 =1 so the statement is true when n=1. Assume statement is true for n=k. That is we assume 1^2+2^2+3^2+...k^2=(k(k+1)(2k+1))/6 Now, we show the statement is true for n=k+1. 1^2+2^2+3^2+...+k^2+(k+1)^2 by induction hypothesis we know that 1^2+2^2+3^2+...+k^2 = (k(k+1)(2k+1))/6 if we add (k+1)^2 to both sides we get: 1^2+2^2+3^2+...+k^2 + (k+1)^2 = (k(k+1)(2k+1))/6 + (k+1)^2 Now we simplify the right hand side. (k(k+1)(2k+1))/6 + (k+1)^2 = ( k(k+1)(2k+1))/6 + 6(k+1)^2/6= ( k(k+1)(2k+1) + 6(k+1)^2 ) /6= [ (k+1)(k(2k+1) + 6(k+1) ) ]/6= [(k+1)(2k^2 +7k +6) ]/6= [(k+1)(2k + 3)(k + 2) ]/6= [(k+1)(2(k+1)+1)((k+1)+1)]/6 this shows the statement is true for n =k+1 so the statement is true for all n by induction.
Permutations and Combinations: If there are 'n' lines on a plane, no two being parallel, and no three or more concurrent, in how many parts will they divide the plane?
I shall begin with an empty plane and populate it one by one with the [math]n[/math] lines.Let [math]P[/math] be the property that no two lines on the plane are parallel, and no three or more concurrent.Let [math]f(n)[/math] be the number of regions in the plane when there are [math]n[/math] lines on the plane. Note that [math]f(1) = 2.[/math]Suppose I have added [math]k[/math] lines (out of [math]n[/math]) to the plane. The number of regions at that point is [math]f(k)[/math]. Since all the [math]n[/math] lines share property [math]P[/math], when we add the [math](k+1)^{th}[/math] line, it will intersect every other line. Hence, it will pass through exactly [math]k+1[/math] regions (one region between every pair of consecutive intersection points and one region at each end of the [math](k+1)^{th}[/math] line). Corresponding to every region that the [math](k+1)^{th}[/math] line passes through, two new regions are created and one old region is deleted. This gives us the following relation between [math]f(k)[/math] and [math]f(k+1)[/math]:[math] f(k+1) = \left(f(k) - (k+1)\right) + 2(k+1) [/math]or[math] f(k+1)= f(k) + k+1[/math]We can solve the recurrence relation with [math]f(1)=2[/math] to get[math]f(n) = \dfrac{n(n+1)}{2} + 1[/math]We could also use induction to prove this formula.We can also get the number of intersection points and the number of edges by slightly modifying the above argument.
Use mathematical induction to prove: 2 is a factor of n^2-n+2?
First, check for n = 0 n² - n + 2 = 0 + 0 + 2 = 2 which is divisible by 2 True for n = 0 Now we show that if statement is true for some n=k, then it is also true for n=k+1. True of n = k → 2 is a factor of k² - k + 2. Therefore, for some integer m: k² - k + 2 = 2m Now we check that statement is true for n=k+1 (k+1)² - (k+1) + 2 = (k² + 2k + 1) - (k + 1) + 2 . . . . . . . . . . . . . = k² + 2k + 2 - k . . . . . . . . . . . . . = (k² - k + 2) + 2k . . . . . . . . . . . . . = 2m + 2k . . . . . . . . . . . . . = 2 (m+k) So 2 is a factor of n² - n + 2 for n = k+1 Therefore when statement is true for n=k it is also true for n=k+1 By mathematical induction, statement is true for all n ≥ 0 -------------------- Note: you can also show that this is true for all integers (negative and positive) Just simply use above method, but show that if statement is true for n=k, then it is also true for n=k-1 The only difference is the step where we check that statement is true for n = k-1: ... Now we check that statement is true for n=k-1 (k-1)² - (k-1) + 2 = (k² - 2k + 1) - (k - 1) + 2 . . . . . . . . . . . . . = k² - 2k - k +1 + 1 + 2 . . . . . . . . . . . . . = (k² - k + 2) - 2k + 2 . . . . . . . . . . . . . = 2m - 2k + 2 . . . . . . . . . . . . . = 2 (m-k+1) So 2 is a factor of n² - n + 2 for n = k-1 Therefore when statement is true for n=k it is also true for n=k-1 By mathematical induction, statement is true for all n ≤ 0 -------------------- Since statement is true for all n ≥ 0 and all n ≤ 0, then it is true for all integers
1)prove:the order of a permutation is the least common multiple of the orders of the cycles which make it up?
(1) Hint. Let p be a permutation that acts on {1,...n} and let d(i) be the size of the cycle of p containing i, i=1,2,...n. Let M be the LCM(d(1),d(2),...,d(n)). (a) Prove that if p^M' = identity, then M' is divisible by d(i) for all i=1,...,n, and hence, must be divisible by M. (b) Prove that p^M = identity. That shows that M is the order of p. (2) (a) Prove it is true for n=1,2. (b) Use induction: Given any p in S_(n+1), we first want to write p as in the form: p = t1.t2.t3...tk.p' where ti's are transpositions of this sorrt, and p' has the property that p'(n+1)=n+1. Then use the induction step to show that p' can be written as transpositions of this sort.
How do I prove the Permutation formula?
By mathematical induction:Let P(n) be the number of permutations of n items.If you add one more item, then you can form P(n)*n permutations by placing your new item in front of every item in all the P(n) permutations, plus n more permutations by placing it at the end of each permutation. So,P(n+1) = P(n)*(n + 1)As an example, here is the case for n = 3 for which P(3) = 6.abcacbbcabaccabcbaIf you add one more item, d, you get:d-abc, a-d-bc, ab-d-c and abc-dd-acb, a-d-cb, ac-d-b and acb-dd-bca, b-d-ca, bc-d-a and bca-dd-bac, b-d-ac, ba-d-c and bac-dd-cab, c-d-ab, ca-d-b and cab-dd-cba, c-d-ba, cb-d-a and cba-d.So P(4) = P(3)*(3+1) = 6*4 = 24 permutations.We know that P(1) = 1 because a single item can have only itself as the single permutation.Hence,P(2) = P(1)*(1 + 1) = 1*2 = 2!P(3) = P(2)*(2 + 1) = 2!*3 = 3!. . . . . .P(n) = P(n - 1)*n = (n - 1)! * n = n!Q.E.D. (Quod erat demonstrandum)
What is proof of the derangement formula in permutation and combination?
Well, that's a nice question I had come across Quora after so many days.That's a bit ingenuine.We will use recurrence relation.Suppose f(n) denotes number of dearrangements of n letter in n boxes. Number the boxes 1,2,3,…,n. And number the letter 1,2,3,..,n. So 1 must not go in 1 and so on.Consider f(n+1). It is the number of dearrangements of n+1 letters in n+1 boxes.Let us consider the box 1 (the first box). Fix 1 in box 2 and 2 in box 1.There are f(n-1) ways to dearrange the remaining letters.Since any of the 2,3,… ,n+1 can come, there are nf(n-1) ways.Now, we twist the story a bit. Let letter 1 does not go in 2. Fix 2 still in 1. What if we readdress the box 2 as 1 for a minute? Yes, it would become dearrangements of n things i.e. f(n) ways. Since 2,3,…,n+1 can occupy the first box, there are nf(n) ways.So, f(n+1) = n( f(n) + f(n-1)) .This recurrence can be solved along with f(1) =0 , f(2) = 1.SO, SOLVING THE RECURRENCE YIELD THE ANSWER TO OUR PROBLEM BRO. !YOU can check thatF(n) = n! (1–1/2!+1/3!-… upto n).Satisfies the recurrence relation.(If you want to actually solve the recurrence, you can use induction. ! )
How do you prove by mathematical induction that [math]2^n > n^3[/math] for all [math]n \geq 10[/math]?
I am going to have fun with this one.Proof by induction that [math]2^n > n^3[/math] for all [math]n \geq 10[/math].First we prove the base case [math]n = 10[/math]:[math]1024 = 2^{10} > 10^3 = 1000 [/math]Check. The base case is proven.Now, the inductive step:Assume [math]2^k > k^3[/math]. We want to show that [math]2^{k+1} > (k+1)^3[/math].Take the derivative of both sides of the inductive hypothesis inequality with respect to [math]k[/math], and now we need to prove that:[math]ln(2)2^k > 3k^2[/math]Well, note that the inequality holds since,[math]ln(2)2^k > 2^k > k^3 > 3k^2[/math]and dividing the rightmost inequality by [math]k^2[/math], [math]k > 3[/math], for [math]n\geq 10[/math], and since we assumed inductive hypothesis was true, we have proven,[math]ln(2)2^k > 3k^2[/math]and because the derivatives of the LHS is always greater than the derivative of the RHS we have shown our inductive step must certainly be true (indeed, we have already proven it for all [math]k_1 >[/math] [math]k[/math]) . Writing that explicitly, we have shown,[math]2^{k+1} > \,[/math]([math]k+1)^3[/math]and that establishes the inductive step. Q.E.D.