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Re: For any positive integer n, the sum of the first n positive integers
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25 Feb 2013, 08:42
Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first  last) / frequency) + 1. You can also multiply by the average at the end to get the sum. In your case it is (((30199)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit? Hope this helps! Ron
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Re: For any positive integer n, the sum of the first n positive integers
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27 Apr 2016, 08:42
gwiz87 wrote: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?
A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150
Hi,
Hoping you can explain this one to me. Although a formula is provided in this problem, we can easily solve it using a different formula: sum = (average)(quantity) Let’s first determine the average. In any set of numbers in an arithmetic sequence, we can determine the average using the formula: (1st number in set + last number in set)/2 Remember, we must average the first even integer in the set and the last even integer in the set. So we have: (100 + 300)/2 = 400/2 = 200 Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive. Two key points to recognize: 1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities. 2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction. quantity = (300 – 100)/2 + 1 quantity = 200/2 + 1 quantity = 101 Finally, we can determine the sum. sum = 200 x 101 sum = 20,200 Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use. Answer is B.
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Re: For any positive integer n, the sum of the first n positive integers
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31 Dec 2017, 17:29
niks18I was confused if the second formula was applied only when my sequence is starting from 1.
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Re: For any positive integer n, the sum of the first n positive integers
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31 Dec 2017, 21:19
adkikani wrote: niks18I was confused if the second formula was applied only when my sequence is starting from 1. Hi adkikaniThe formula is a simple derivation of AP series that starts from 1 and is consecutiveSo the Sum of AP series is \(S_n=\frac{n[2a+(n1)d]}{2}=\frac{n}{2}\)[First term + Last Term] if the sequence starts from 1 and is consecutive so your First term=1 and Last Term=number of terms in the series (because sequence is consecutive) so \(S_n=\frac{n(1+n)}{2}\) Let's take a simple example. if I have to find sum of numbers from 6 to 10, then I can do that by first finding sum of all Numbers from 1 to 10 and then from this sum subtract sum of numbers from 1 to 5. In the number line this can be represented as 1...2...3...4...5... 6...7...8...9...10. Here First term =1 and last term=10=number of terms in the sequence As you can clearly see to find sum of numbers from 6 to 10, we simply need to remove sum of numbers from 1 to 5 out of the total sum i.e from 1 to 10. Now in this question we need to find sum of number from 50 to 150 so it will be Total sum of numbers from 1 to 150 less sum of numbers from 1 to 49



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Re: For any positive integer n, the sum of the first n positive integers
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19 Apr 2018, 13:46
gwiz87 wrote: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?
A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150 Here's one approach. We want 100+102+104+....298+300 This equals 2(50+51+52+...+149+150) From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150  50 + 1 = 101) To evaluate 2(50+51+52+...+149+150), let's add values in pairs: ....50 + 51 + 52 +...+ 149 + 150 + 150+ 149+ 148+...+ 51 + 50...200+ 200+ 200+...+ 200 + 200 How many 200's do we have in the new sum? There are 101 altogether. 101 x 200 = 20,200 Answer: B Approach #2: From my last post, we can see that we have 101 even integers from 100 to 300 inclusive. Since the values in the set are equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200 So, we have 101 integers, whose average value is 200. So, the sum of all 101 integers = (101)(200) = 20,200 = B Approach #3: Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150) From here, we'll evaluate the sum 50+51+52+...+149+150, and then double it. Important: notice that 50+51+.....149+150 = (sum of 1 to 150)  (sum of 1 to 49) Now we use the given formula: sum of 1 to 150 = 150(151)/2 = 11,325sum of 1 to 49 = 49(50)/2 = 1,225So, sum of 50 to 150 = 11,325  1,225 = 10,100So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200 Answer: B Cheers, Brent
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Re: For any positive integer n, the sum of the first n positive integers
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15 Nov 2018, 23:46
We are given the formula: \(∑_1^n \frac{n(n+1)}{2}\) We're asked to find sum of all the even numbers between \(99\) and \(301\).... \(100+102+104+...+296+298+300\) We can factor out \(2\) from this sum: \(2 \times (50+51+52+...+148+149+150)\) In the brackets we have sum of numbers from \(50\) to \(150\). We can use the given formula for the sum of \(1\) to \(150\) and subtract the sum of \(1\) to \(49\) to find the total of this sum. \(2 \times (\frac{150 \times 151}{2}\frac{49 \times 50}{2})\) \(=(150)(151) – (49)(50)\) \(=22,650 – 2,450\) \(=20,200\) The final answer is .



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Re: For any positive integer n, the sum of the first n positive integers
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09 Apr 2019, 14:56
CharmWithSubstance wrote: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?
A. 10,100 B. 20,200 C. 22,650 D. 40,200 E. 45,150 So I know, that we must use the first and last digits of even numbers, if we are calculating for a sum of evens.. But I tried it this way, and still got the answer.. Would like to know, if it's an acceptable method, or its wrong, and I shouldn't use it ? if we use first and last terms as given, i.e. First term= 99 and last term = 301 Subtract these two numbers, to get the total number of terms.. 30199 = 202 Note this includes odd and evens both for even, we must divide it by 2. therefore, we get a 202/2 = 101 terms... AS you can see, nothing is added or subtracted for inclusive, exclusive.. now, 101 * avg of terms, will give us the sum. avg = 301+99/2 = 400/2 = 200 101*200 = 20,200 option B Please correct me, if you think this is a wrong approach and we should not use it




Re: For any positive integer n, the sum of the first n positive integers
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