Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. x→−3lim x2 + 2x − 3x2 − 9. - Mathematics . step-by-step.. ((2n-1)!)/((2n+1)!) = 1/((2n+1)(2n)) Remember that: n! =n(n-1)(n-2)1 And so (2n+1)! =(2n+1)(2n)(2n-1)(2n-2) 1 Solve your math problems using our free math solver with step-by-step solutions. Base step (n = 0 n = 0 ): S(0) S ( 0) says that 20 = 21 − 1 2 0 = 2 1 − 1, which is true. View Solution. Follow answered Jan 16, 2018 at 0:48. $$ Remark: I suggest this proof since the plain inductive proof of your statement has been given in many answers. My Notebook, the Symbolab way. series 1/2^n. discrete math. Voici le corrigé de la démonstration par récurrence à faire : 2^n>n² pour n>4.2, which arrived with the Journal app earlier this month. It is easy to apply the formula when the value of n is known. n = 1 → LH S = 12 = 1. The United States was one of 10 countries voting against the resolution. Reduce the expression by cancelling the common factors. These terms ensure that each object is only counted once and that the order in which they are chosen does not matter.. Prove that, 2n+1 < 2 n, for all natural number n ≥ 3. I have to prove that $1^2 + 3^2 + 5^2 + + (2n-1)^2 = \frac{n(2n-1)(2n+1))}{3}$ So first I did the base case which would be $1$. Possible Duplicate: Proof the inequality n! ≥2n by induction Prove by induction that n! >2n for all integers n ≥ 4.2. Which correspond to the formula $2^n - 1$ (predicted by the algorithm) So I was trying to prove that the sum of this series will result in $2^n - 1$ but did not succeed.. Dec 20, 2022 - Air Force 67 vs. holds real estate brokerage licenses in multiple provinces. -2^ (-2n) + 2^ (-2n) = 1/4^n + 1/4^n = 2/2^2n = 2^ (1 - 2n) Answer: D.1. Differentiation. Show that the number is $𝑛(𝑛 + 1)/2$ by considering the number of $2$-lists $(𝑎, 𝑏)$ in which $𝑎 > 𝑏$ or $𝑎 < 𝑏$. Colorado 65 Russian President Vladimir Putin makes a TV address after Yevgeny Prigozhin's attempted mutiny on Saturday. 1 $\begingroup$ You state that n+1<2n. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.1.8k 4 30 53. Question15 Prove the following by using the principle of mathematical induction for all n N: 12 + 32 + 52 + . December 13, 2023 at 1:59 a. Differentiation.H. Matrix. I'm preparing to an exam and trying to solve an = 2an−1 −an−2 +2n a n = 2 a n − 1 − a n − 2 + 2 n, where a0 = 0 a 0 = 0 and a1 = 1 a 1 = 1. step is $\;(2(n+1))^2\;$ , and Prove that 1 + 2 + 22 + + 2n = 2n+1 - 1 for All N ∈ N . $$ All the terms are positive; observe that $$ \binom{n}{1} = n, \quad \binom{n}{n-1} = n. so we need to find the lowest natural number which satisfies our assumption that is 3. Enter a problem Cooking Calculators. Related Symbolab blog posts.. Alternate: $$ n + 1 < 2n < 2 \cdot 2^n = 2^{n+1}, $$ as desired. Hence, the max number you can represent is 2^3-1=7. The inductive step can be proved as follows. 22n(2n+1) −2( 2n(n+1)) = n(2n+1)− n(n+ 1) = n2. n2 − 2⋅n⋅1+12 n 2 - 2 ⋅ n ⋅ 1 + 1 2 This question already has answers here : Prove that n <2n n < 2 n for all natural numbers n n. Second part: 22n > (2n n). General Assembly voted Dec. Then adding up the sizes of each subset gives $0+1+1+2 = 4$. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. In summary, the homework statement states that 2n ≤ 2^n holds for all positive integers n. This can be done by substituting n+1 into the original statement and simplifying until it matches the statement for n. I am trying to learn the proper way to prove things like this but everything I read confuses me even more. For example, the possible subsets of $\{1,2\}$ are $\{\},\{1\},\{2\},\{1,2\}$. Induction step (S(k) → S(k + 1) S ( k) → S ( k + 1) ): Fix some k ≥ 0 k ≥ 0 and suppose that.N2# eunevA nialpmahC S 6584 .2 = ndnl ∞→nmil . $\endgroup$ – BlueRaja - Danny Pflughoeft For all n ϵ N, 3. 2 k $\begingroup$ @gaurav: At that link you will find other methods that can be applied here. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let P (n) be the statement that 1² + 2² + · · · + n² = n(n + 1)(2n + 1)/6 for the positive integer n.+(2n 1)2 = (n(2n 1)(2n + 1))/3 For n = 1, L. Best answer. Tap for more steps a = 2− 1 n a = 2 - 1 n Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework … Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(n-1\right)^{2}. I must show that it converges to 2. Since contains both numbers and variables, there are two steps to find the LCM. Share. Aug 23, 2011 at 10:01 2 (n + 1)3 −n3 = 3n2 + 3n + 1 - so it is clear that the n2 terms can be added (with some lower-order terms attached) by adding the differences of cubes, giving a leading term in n3. 2n = 2⋅n ⋅1 2 n = 2 ⋅ n ⋅ 1 Rewrite the polynomial. So that makes 2 k + 1 + 2 < 2 k + 2 and since it was assumed k ≥ 3 we also know that 2 < 2 k. Arithmetic.+ 2^n.N. Let k k be the smallest number that k(2n − 1) k ( 2 n − 1) has at most n − 1 n − 1 ones in binary expansion. a) What is the statement P (1)? b) Show that P (1) is true, completing the basis step of the proof. \sum_ {k=1}^n (2k-1) = 2\sum_ {k=1}^n k simplify \frac{(n+1)^{2}}{(n+2)^{2}} en. Transcript. Simultaneous equation. Verified by Toppr. The base n = 1 is trivial. In this case, the geometric progression summation formula will help us. However, to prove this formally, the author needs to show that k+1 holds for all positive integers n. となる。メルセンヌ数は2進法表記で n 桁の 11⋯11 、すなわちレピュニットとなる。. The technique of Courant can also be used on an = {P This proof uses the triangulation definition of Catalan numbers to establish a relation between C n and C n+1. probability. - Mathematics Stack Exchange dn = (1 + (2/n))n converge or diverge and find the limit? Ask Question Asked 8 years, 8 months ago Modified 8 years, 8 months ago Viewed 29k times 2 I know the answer is e2 and I'd like to use L'Hopital's rule because this is an indeterminate form. Démonstration par récurrence 2^n>n². Given an integer N, the task is to find the sum of series 2 0 + 2 1 + 2 2 + 2 3 + ….0k points) principle of mathematical induction I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list: Prove, using mathematical induction, that $2^n > n^2$ for all integer n greater than $4$. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.53444°E. The number k 2 < k k 2 < k generates at most n − 1 n − 1 ones in k 2(2n − 1) k 2 ( 2 n − 1) as well, contradiction. lndn = ln((1 + 2 n)n) = n ln(1 + 2 n) = ln(1 + 2 n) 1 n. Math notebooks have been around for hundreds of years.
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1 is descended from BA. Tap for more steps a = 2− 1 n a = 2 - 1 n Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Share 2N, 2N+1, 2N+2 redundancy. Tap for more steps n2(2n) +n2 ⋅1+n(2n)+n⋅1 n 2 ( 2 n) + n 2 ⋅ 1 + n ( 2 n) + n ⋅ 1. Syllabus.Tech from Indian Institute of Technology, Kanpur. Prove by induction that (1 x 1!)+ (2 x 2!)++ (n x n)= (n+1)!-1. For math, science, nutrition, history, geography, engineering, mathematics Learn more. It makes everything more concise and easier to manipulate: ∑i=1k+1 i ⋅ i! =∑i Hint only: For n ≥ 3 you have n2 > 2n + 1 (this should not be hard to see) so if n2 < 2n then consider 2n + 1 = 2 ⋅ 2n > 2n2 > n2 + 2n + 1 = (n + 1)2. Follow answered Oct 21, 2013 at 15:57. 3 Answers. In general, (2n)! is enormously larger than n!. A quick recap of redundancy levels includes key terminology such as N, N+1, N+2, 2N, 2N+1, 2N+2, 3N/2. Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their The proof by induction for 2^n < n can be done by first proving the base case, which is usually n = 1. Even more succinctly, the sum can be written as. 22n+1−n2 2 2 n + 1 - n 2. answered Aug 25, 2012 at 3:10. Limits. N+1, N+2, 2N, 2N+1: A redundancy model to meet the needs of every business. For math, science, nutrition, history Algebra. Improve this answer. Simply 2n−1 +2n−1 = 2 ⋅2n−1 =2n which works for the base 2 - for base three you'd need to add three times 3n−1.3 Answers Sorted by: 1 In the induction hypothesis, it was assumed that 2 k + 1 < 2 k, ∀ k ≥ 3, So when you have 2 k + 1 + 2 you can just sub in the 2 k for 2 k + 1 and make it an inequality. My Notebook, the Symbolab way. Therefore the series ∑∞n = m n! ( 2n)! converges. The first + the last; the second + the one before last. Try to make pairs of numbers from the set. [3] Tower 1, at 302 metres (991 feet) tall with 65 floors, is the ninth-tallest building in The Moscow International Business Center ( MIBC ), [a] also known as Moscow-City, [b] is an under-construction commercial development in Moscow, the capital of Russia. It means n-1 + 1; n-2 + 2. Let P (n) be the statement that 1² + 2² + · · · + n² = n (n + 1) (2n + 1)/6 for the positive integer n. dxd (x − 5)(3x2 − 2) Integration. holds and we need to prove: (k + 1)! ⋅ 2k + 1 ≤ (k + 2)k + 1. Limits. However to start the … answered Mar 14, 2015 at 16:52. For n ≥ 0 n ≥ 0, let S(n) S ( n) denote the statement. For the inductive step, assume that for some n ≥ 5, that n2 < 2n. Advertisement Hint: consider the the set of all subsets of $\{1,2,\dots,n\}$ (of which there are $2^n$) and try to find the total sum of the sizes of the subsets in two different ways. View Solution. Radical equations are equations involving radicals of any order. Cite. First part: 2n < (2n n). So there are 6 possible combinations with 4 items. Detailed step by step solution for ( (2 (n+1))!)/ ( (2n)!) Popular Problems Algebra Factor n^2-2n+1 n2 − 2n + 1 n 2 - 2 n + 1 Rewrite 1 1 as 12 1 2. However as n is basically infinite above 2 I don't think this is right. Clearly if I take x = 1 2 x = 1 2 , the series is ∑∞ n=0 n 2n ∑ n = 0 ∞ n 2 n. The principle of mathematical induction can be extended as follows. In the induction hypothesis, it was assumed that 2 k + 1 < 2 k, ∀ k ≥ 3, So when you have 2 k + 1 + 2 you can just sub in the 2 k for … Let P(n) be the given statement, i. Assume that … Simplify the right side. A list Pm, > Pm + 1, ⋯ of propositions is true provided (i) Pm is true, (ii) > Pn + 1 is true whenever Pn is true and n ≥ m. A naive approach is to calculate the sum is to add every power of 2 from 0 to n. And since you are adding two numbers together, there are only (n-1)/2 pairs that can be made from (n-1) numbers.2. Example 1 For all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n(n+1)(2n+1))/6 Let P(n) : 12 + 22 + 32 + 42 + …. Explanation: using the method of proof by induction. Thus, the contrapositive of the original statement is as follows: n = b* (2^k), where b is a positive odd number ==> 2^n + 1 is composite. Unfortunately, your claim is false. Math notebooks have been around for hundreds of years. Simultaneous equation. A power of two is a number of the form 2n where n is an integer, that is, the result of exponentiation with number two as the base and integer n as the exponent . Integration. Algebra Simplify (n-1) (2n-2) (n − 1) (2n − 2) ( n - 1) ( 2 n - 2) Expand (n−1)(2n− 2) ( n - 1) ( 2 n - 2) using the FOIL Method. Matrix. $\begingroup$ Because the rule is: "Begin with some natural number $\;n\;$ when one is added, and end with twice that number $\;n\;$ " Thus, when in the inductive step with begin with $\;n+1\;$ and add one to it, we must end with $\;2(n+1)\;$ But we sum consecutive naturals, so if the last one in the first step is $\;(2n)^2\;$ , the last one in the ind.D. The project occupies an area of 60 hectares, [1] and is located just east of the Third Ring Road at the western edge of the Presnensky District in the Central Administrative Okrug.2. 12 to demand a humanitarian cease-fire in Gaza. Reduce the expression by cancelling the common factors. However to start the induction you need something greater than three. I'd say, that if \frac{n(n+1)}{2} is som of n numbers, then \frac{(n-1)n}{2} is the sum of n-1 numbers, do you agree?.(2n-1)} 2 n.459 x ≈ 3. 16. The proof itself can be done easily with induction, I ass This assumption is called the inductive assumption or the inductive hypothesis. 2^ (2n) can be expressed as … The sum of the first n n even integers is 2 2 times the sum of the first n n integers, so putting this all together gives. I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list: Prove, using mathematical induction, that $2^n > n^2$ for all integer n greater tha We have proved the contrapositive, so the original statement is true. to prove n+1! > 2n. We can do this 6 and $(1)$ follows from $(2)$ and the sum formula for the arithmetic progression $$1+2+3+\ldots +n=\frac{\left( n+1\right) n}{2}.As a base case, if n = 5, then we have that 52 = 25 < 32 = 25, so the claim holds. As you enter, you'll be greeted by the spacious and open floor plan, creating a welcoming ambiance with a licenses in multiple states. Tap for more steps 2n3 + 3n2 +n 2 n 3 + 3 n 2 + n. 0 : 000 1 : 001 2 : 010 3 : 011 4 : 100 5 : 101 6 : 110 7 : 111 Anything beyond that requires more than 3 digits. Add n n and n n. This question can be solved by method of induction. Let us learn to evaluate the sum of squares for larger sums. Bernard's answer highlights the key algebraic step, but I thought I might mention something that I have found useful when dealing with induction problems: whenever you have an induction problem like this that involves a sum, rewrite the sum using -notation. M n = 2 n − 1 が素数ならば n もまた素数であるが、逆は成立しない (M 11 = 2047 = 23 × 89)。素数であるメルセンヌ数をメルセンヌ素数（メルセンヌそすう、英: Mersenne prime ）という。 なお、「メルセンヌ数」という 4 Answers. Differentiation. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. \frac {2n (2n+1)}2 - 2\left ( \frac {n (n+1)}2 \right) = n (2n+1)-n (n+1) = n^2. 7,606 5 5 gold badges 28 28 silver badges 64 64 bronze badges $\endgroup$ 2. Applying squeeze theorem on (1) we get (logP(n)) / Q(n) → 0 and hence an → 1. 2N simply means that there is twice the amount of required resources/capacity … Sum of the series 2^0 + 2^1 + 2^2 +…. ∙ prove true for some value, say n = 1. Limits. Share. Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.1 is available now for iPhone XS and According to Guns N' Roses bassist Duff McKagan, though, there's a good reason for the band's decision to play roughly three-and-a-half-hour live sets. Now this means that the induction step "works" when ever n ≥ 3. Now this means that the induction step "works" when ever n ≥ 3. We observe that P(n) is true, since. Add a comment. You write down problems, solutions and notes to go back Read More. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.S = (1(2 1 1)(2 1+ 1))/3 = (1(2 1) (2 + 1))/3 = (1 1 3)/3 = 1 Hence L. For example, in Preview Activity 4. $\endgroup$ - BlueRaja - Danny Pflughoeft answered Mar 14, 2015 at 16:52. Thus, you can extend this to any n and say that you can express integers in the range [0,2^n -1]. Given an integer N, the task is to find the sum of series 2 0 + 2 1 + 2 2 + 2 3 + …. Q 3. If we rewrite $2^{2k}-1$ in binary, we get the number $$2^{2k}-1=\underset{\text{$2k$-times}}{\underbrace{111111\dots11}}$$ consisting of $2k$ ones. Prove that:(2n)!/n! = { 1*3*5. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Fredrik Meyer. In a context where only integers are considered, n is restricted to non-negative values, [1] so there are 1, 2, and 2 multiplied by itself a certain number of times.$$ (Adapted from Proof 1 in this answer to the question Prove that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?; see the above Theo Buehler's comment. Human Rights Office said it was calling for an investigation of what transpired during the raid, citing allegations from medical staff that patients had died because of the conditions in A man who was studying for a Ph. André Nicolas André Nicolas. Je ne comprends pas tout le raisonnement, comment en est on arrivé à la dernière ligne ? Solve for n 1/(n^2)+1/n=1/(2n^2) Step 1.. Simplify and combine like terms. n2 − 12 n 2 - 1 2. And since you are adding two numbers together, there are only (n-1)/2 pairs that can be made from (n-1) numbers. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. $\begingroup$ Another way to say this is that each subset can be tagged with a binary number constructed by using $ \ n \ $ digits and writing "0" or "1" at each digit according to whether the $ \ k^{th} \ $ element is in the subset. So now we have 2 k + 1 + 2 < 2 k + 2 < 2 k + 2 k = 2 k + 1 . Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Redundancy can be broken down into several different levels. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Method 1: You can take a graphical approach to this problem: It can be seen that the graphs meet at (0, 1), 2x 2 x is greater until they intersect when x ≈ 3.459 x ≈ 3. Applying the intuitive understanding of division as repeated subtraction, we can plot 12 on a numberline, and then since we are dividing by 2, we count backwards by 2 until we reach 0. Natural Language; Math Input; Extended Keyboard Examples Upload Random. 3. 7. Find the LCD of the terms in the equation. There are (4n + 2)C n such marked triangulations for a given base. I present my two favorite proofs: one because of its simplicity, and one because I came up with it on my own (that is, before seeing others do it - it's known). Organizations are continuing to embrace digital transformation to support operations and drive business growth. Step 1. Tap for more steps (n2 + n)(2n+1) ( n 2 + n) ( 2 n + 1) Expand (n2 +n)(2n+1) ( n 2 + n) ( 2 n + 1) using the FOIL Method. An efficient approach is to find the 2^ (n+1) and subtract 1 from it since we know that 2^n can be written as: Feeling lost O (2^ (n+1)) is the same as O (2 * 2^n), and you can always pull out constant factors, so it is the same as O (2^n).M. Simply 2n−1 +2n−1 = 2 ⋅2n−1 =2n which works for the base 2 - for base three you'd need to add three times 3n−1. However, constant factors are the only thing you can pull out. 2. We will start by introducing the geometric progression summation formula: $$\sum_{i=a}^b c^i = \frac{c^{b-a+1}-1}{c-1}\cdot c^{a}$$ Finding the sum of series $\sum_{i=1}^{n}i\cdot b^{i}$ is still an unresolved problem, but we can very often transform an unresolved problem to an already solved problem. If you don't like it, I won't be at all offended if you revert! @NicholasR. c) What is the The proof I am dealing with is worded exactly as follows: Prove $\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$. One of those is an even number, so we've added at least one factor of 2.+ n2 = (𝑛(𝑛 + 1)(2𝑛 + 1 1.N. Concept Notes & Videos 127. This is proven easily enough by splitting it up into two parts and then proving each part by induction. N refers to the minimum number of resources (amount) required to operate an IT system.ot lauqe si 1 n3 2n2 n21 n2ncarf elytsyalpsidytfniotnmil elytsyalpsid:dnah_gnitirw: noitseuq ruoy ot rewsna na teg ot:2_pu_tniop:ereh kcilC . [2] The first ten a. = R. Q4. Integration. Tap for more steps 2n+1−n2+n 2 n + 1 - n 2 + n. Plugging 4 into the equation we get 4(4-1)/2 = 12/2 = 6. The 1,100 Square Feet unit is a 1 bed, 1 bath apartment unit.
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O (n^2) < O (2^n) means there is some N such that for all n > N, a*n^2 < b*2^n, for any choice of positive constants a and b. [duplicate] (12 answers) Closed 5 years ago.4, 10 Find the sum to n terms of the series whose nth terms is given by (2n 1)2 Given an = (2n 1)2 = (2n)2 + (1)2 2 (2n) (1) = 4n2 + 1 4n = 4n2 4n + 1 Sum of n terms is = 4 ( (n (n+1) (2n+1))/6) 4 (n (n+1)/2) + n = n ("4" (n (n+1) (2n+1))/6 " 4". Then, the inductive step involves showing that if the statement is true for n, it is also true for n+1.
$\begingroup$ Another way to say this is that each subset can be tagged with a binary number constructed by using $ \ n \ $ digits and writing "0" or "1" at each digit according to whether the $ \ k^{th} \ $ element is in the subset.
Re: If n is a positive integer, then (-2^n)^ {-2} + (2^ {-n})^2 is equal to [ #permalink ] Mon Mar 28, 2016 9:57 am. The result is always n. Integration.H. Even more succinctly, the sum can be written as.Peterson Thanks for the edit. But this isn't true for n=0. It means n-1 + 1; n-2 + 2. Add a comment. (integrate 1/2^n from n = 1 to xi) / (sum 1/2^n from n = 1 to xi) plot 1/2^n. iOS 17. Follow edited Aug 25, 2012 at 12:14.
7. CBSE Commerce (English Medium) Class 11. However, to prove this formally, the author needs to show that k+1 holds for all positive integers n.
3. Consider the number k+1 2 k + 1 2.1, one of the open sentences P(n) was.
2n+1 (2n)n−1 2 n + 1 ( 2 n) n - 1. This proves your product must be positive.5^2n + 1 + 23^n + 1 is divisible by asked Sep 4, 2020 in Mathematical Induction by Shyam01 ( 51. Ex 9. An efficient approach is to find the 2^ (n+1) and subtract 1 from it since we know that 2^n can be written as: Feeling lost
O (2^ (n+1)) is the same as O (2 * 2^n), and you can always pull out constant factors, so it is the same as O (2^n). For a simple example, let's consider a server in a data center that has ten servers with an additional ten servers that act as
Sum of the series 2^0 + 2^1 + 2^2 +…. The smallest counterexample is as can be seen on the sequence. Limits. Il suffit donc de prouver que n²>2n+1 pour n>4. We can do this 6
Davneet Singh has done his B. Thanks for any help!
Certain things are not transparent when expressed in symbols. The factor 1/3 attached to the n3 term is also obvious from this observation. 494 Lee St #2N, Des Plaines, IL 60016 is an apartment unit listed for rent at $1,490 /mo.
prove: $2n+1\le 2^n$ by induction. He has been teaching from the past 13 years.
1. ( n − 1) + ( n − 2) ⋯ ( n − k) = n + n + ⋯ + n ⏟ k copies − ( 1 + 2 + ⋯ k) = n k − k 2 ( k + 1) I edited your post to put the "underbrace" there; I think it makes this sort of thing more readable.