Prove by Mathematical Induction: n factorial <= 2^(n-1)?
You might have mistyped “<=.” The inequality you mentioned should be n!≥2^(n-1) for all n in N. …(*) Proof of (*): We will prove it by induction. Basis Case: The case n=1 : (*) is obviously true. Inductive Step: We assume (*) is true for n=k (k≥1). We then have k!≥2^(k-1). Multiplying both sides by k+1(≥2), we have (k+1)*k!≥(k+1)*2^(k-1). Since k+1≥2, (k+1)*k!≥(k+1)*2^(k-1)≥2*2^(k-1). Hence, we have (k+1)!≥2^((k+1)-1). It means (*) is true for n=k+1. We complete our induction, and have shown that (*) is true.
How can I prove by principle of mathematical induction that for all integers [math]n \ge 1[/math] [math]1\cdot 1! +2 \cdot 2! +3 \cdot 3!+ \ldots +n\cdot n! =(n+1)!-1?[/math]
Oooh, an A2A.Someone already went all Yahoo! Answers on this, but I'll try to give the answer that I would consider most helpful.I hope you at least know that the two main parts of your proof are:Prove that the base case holds. If you're proving that something is true for all [math]n \geq blah[/math], then you want to put [math]blah[/math] in the place of [math]n[/math] and prove that the thing is true in that case. (For example, if I had been asked to prove by induction that [math]1+2+\cdots +n = \frac{n(n+1)}{2}[/math] for [math]n \geq 1[/math], then for my base case I want to confirm that [math]1 = \frac{1(1+1)}{2}[/math].)Prove that the induction step works. In this case you're assuming that the thing we're trying to prove works for a given value of [math]n[/math] in the given domain, and showing that it must then work for the next value of [math]n[/math]. What I like to do for this part is pick some new letter, like [math]k[/math], to represent the value that we're assuming works. Again, if I'm asked to prove that [math]1+2+\cdots +n = \frac{n(n+1)}{2}[/math] for [math]n \geq 1[/math], then I'll say: "Assume that [math]1+2+\cdots +k = \frac{k(k+1)}{2}[/math] for some [math]k \geq 1[/math]. [I wouldn't italicize "some" in the proof; I'm just emphasizing it here so you don't miss it.] We will show that [math]1+2+\cdots +k + (k+1) = \frac{(k+1)(k+1+1)}{2}[/math]." Note that I deliberately wrote the sum [math]1+2+\cdots +(k+1)[/math] in such a way that I could separate the [math]1+2+\cdots +k[/math] part from the last term, because I know I can replace that part of the sum with something helpful from what I'd assumed:[math]1+2+\cdots +k+(k+1)=\frac{k(k+1)}{2} + (k+1)[/math][math]=\frac{k(k+1)}{2} + \frac{2(k+1)}{2}[/math]and so on.The general approach you'd want to take with your problem is similar.
Mathematical induction problem involving factorials?
A proof by induction would work like this: First check that it's true for the first case, so check n=1: Left Side = 1(1!) = 1 Right Side = (1+1)! - 1 = 2 - 1 = 1 Since they're equal, we know the equation is true for n=1. Now assume we already know that equality holds for n=k, so 1(1!) + 2(2!) + ... + k(k!) = (k+1)! - 1 Then 1(1!) + 2(2!) + ... + k(k!) + (k+1)( (k+1)! ) = ( (k+1)! - 1 ) + (k+1)( (k+1)! ) = (1 + (k+1) ) (k+1)! - 1 = (k+2)(k+1)! - 1 = (k+2)! - 1 = ( (k+1) + 1 ) - 1 which shows that equality holds when n=k+1. By induction, it follows that (1)(1!) + (2)(2!) + (3)(3!) + ... + (n)(n!) = (n+1)! - 1 is true for all integers n > 0. (n=0 works too, but you have to realize the left-hand side as an empty sum) ------- To answer the second part, note that since the sum equals (n+1)! - 1, it will exceed 10^9 when (n+1)! - 1 > 10^9 or when (n+1)! > 10^9 + 1 Just consider some factorials. We eventually get 12! = 479001600 which is less than 10^9 + 1 13! = 6227020800 which is greater than 10^9 + 1. Therefore the minimum number of terms occurs when n+1 = 13, so n=12. ------ As a sidenote, there is also a nice combinatorial proof of the given equality. Note that the right-hand side, (n+1)! - 1 counts the number of permutations of 1, 2, ... , n+1 other than the identity (which leaves all the elements fixed). Now just realize that you can partition these permutations based on the first element that isn't fixed. Count each of the sets in your partition.
Math-factorial inequality?
In truth, for this one it's best to plug in some numbers, though it's easier to solve it in the form (n+1)!>1/.001, i.e., (n+1)!>1000. You get these from multiplying both sides by (n+1)! and dividing by .001, and reading it right to left. Now pull out your calculator and start finding some factorials. When you get to one which is greater than 1000, subtract one from it (because you found n+1), and every n from there on up works. You could start multiplying consecutive numbers if you don't have a calculator with factorials. 1!=1 2!=2*1!=2 3!=3*2!=6 4!=4*3!=24 etc.
What does n! mean in Mathematics?
I was doing a math research paper on Pascal's Triangle and Binomial Expansion and I came across the symbol n!. Could someone please kindly explain what this symbol means and what it has to do with Pascal's Triangle and Binomial Expansion? And please don't tell me to do my own homework, because I can't find what this means, and it may be very important to my paper. Thank you.
Please help! Write a program that estimates the value of the mathematical constant 'e' by using the formula?
Write a program with Programming C that estimates the value of the mathematical constant 'e' by using the formula: e=1 + 1/1! + 1/2! + 1/3! + 1/(n-1)! + 1/n! 'e' should be evaluated as a real value. Evaluate factorial also as real, use n = 10 or more to get an accurate answer. This is what I have so far but my source code keeps failing. #include "stdafx.h" #include "stdio.h" int main(){ int num = 3; int factorial; double e = 1.0; printf("Enter a value for num",); for(int i=0;i