Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{17}=\frac{17\times18}{2}=17\times9=153\). Hence, option C is correct. A value such as \(148\) does not result from the correct calculation of \(\frac{17\times18}{2}\). Exam tip: For the sum of the first \(n\) natural numbers, directly use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=50\), so \(\frac{50\times51}{2}=25\times51=1275\). Therefore, the correct answer is 1275. Note that 1225 equals \(\frac{49\times50}{2}\), the sum from 1 to 49. Exam tip: For consecutive natural numbers starting from 1, use \(n(n+1)/2\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=24\), we get \(\frac{24\times25}{2}=12\times25=300\). Therefore, 300 is the correct option. Although 288 is close, it is not obtained from the formula. Exam tip: In such questions, use \(n(n+1)/2\) and divide the even factor by 2 first.

(S_{15}=120) because (\frac{15\times16}{2}=120). In such questions, verify the answer back in the formula.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{40}=\frac{40\times41}{2}=20\times41=820\). Hence, 820 is correct. An answer such as 800 may result from incorrectly using 40 in place of \(n+1\). Exam tip: Write \(n(n+1)/2\) before substituting the value of \(n\).

This is the sum of natural numbers from 1 to 13. Using \(1+2+\cdots+n=\frac{n(n+1)}{2}\), with \(n=13\), we get \(\frac{13\times14}{2}=91\). Hence, 91 is correct. Note that 78 is the sum from 1 to 12, so it is a close but incorrect option. Exam tip: identify the last term as \(n\) and apply \(\frac{n(n+1)}{2}\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=21\), we get \(\frac{21\times22}{2}=21\times11=231\). Hence, option B is correct. \(210\) is the sum from \(1\) to \(20\), so it is a close but incorrect choice. Exam tip: identify \(n\) and \(n+1\) carefully before applying the formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=35\), so \(\frac{35\times36}{2}=35\times18=630\). Therefore, the correct answer is 630. A value such as 610 may result from an error in calculation. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The governing concept is the triangular-number identity 1 + 2 + 3 + ... + n = n(n + 1)/2. Equating this to 210 gives n(n + 1)/2 = 210, and therefore n(n + 1) = 420. For n = 20, the product is 20 × 21 = 420, so the sum is 20 × 21/2 = 210. Thus option C is correct. A neighboring check gives S_19 = 19 × 20/2 = 190 and S_21 = 21 × 22/2 = 231, so 210 lies exactly between these at n = 20. Since triangular sums increase strictly for positive n, no other option can satisfy the equation. The ellipsis means that all consecutive integers from 1 to n are included.

The sum of all numbers is (S_{23}=\frac{23\times24}{2}=276). When numbers start consecutively from (1), use (S_n).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=28\), we get \(\frac{28\times29}{2}=14\times29=406\). Although 396 is close, it is not obtained by correctly dividing \(28\times29\) by 2. Exam tip: Divide the even factor by 2 first to calculate quickly.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). For example, when n = 4, \(1+2+3+4=10\), and the formula gives \(\frac{4\times5}{2}=10\). \(\frac{n(n-1)}{2}\) relates to the sum up to \(n-1\). Exam tip: check for the \(n+1\) factor.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=34\), so \(\frac{34\times35}{2}=17\times35=595\). Therefore, 595 is correct. A value such as 585 may result from an error while multiplying or dividing by 2. Exam tip: when the last term is \(n\), directly use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=36\), we get \(\frac{36\times37}{2}=18\times37=666\). Therefore, 666 is correct. A value such as 676 can result from an arithmetic error. Exam tip: always use the next number, \(n+1\), in the formula.

The roll numbers form the consecutive sequence 1, 2, 3, ..., 31, so they are the first 31 natural numbers. Apply S_n = n(n + 1)/2 with n = 31: S_31 = 31 × 32/2 = 31 × 16 = 496. Therefore option B is correct. Pairing provides an independent check: 1 + 31, 2 + 30, 3 + 29, and so on each total 32. There are 15 such pairs, with the middle number 16 left unpaired, giving 15 × 32 + 16 = 496. The other choices differ by 10 or 20 and can result from an arithmetic error, forgetting the middle term, or using an incorrect number of pairs.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=45\), we get \(\frac{45\times46}{2}=45\times23=1035\). Hence, 1035 is correct. An option such as 1025 may result from an incorrect addition or multiplication. Exam tip: in this formula, always use \(n+1\) along with \(n\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=325\), so \(n(n+1)=650\). Since \(25\times 26=650\), \(n=25\). For \(n=24\), the sum is \(\frac{24\times25}{2}=300\), not 325. Exam tip: check for two consecutive factors of twice the given sum.

The sum of the first 27 natural numbers is found using \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{27}=\frac{27\times28}{2}=27\times14=378\). Therefore, 378 is correct. A value such as 364 can result from an arithmetic error while multiplying or dividing. Exam tip: Cancel the even factor with 2 before multiplying.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=33\), we get \(\frac{33\times34}{2}=33\times17=561\). Therefore, option B is correct. A value such as 551 would result from an arithmetic error. Exam tip: always use the next number, \(n+1\), in the formula.

Total stars are (S_{18}=\frac{18\times19}{2}=171). Increasing row counts form the sum of natural numbers.

(S_{30}=\frac{30\times31}{2}=465), so (n=30). Substituting options in the formula can give a quick answer.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=100\), we get \(\frac{100\times101}{2}=5050\). The value \(5000\) results from incorrectly using \(100\times100/2\) and missing the \(n+1=101\) factor. Exam tip: For \(1+2+\cdots+n\), apply \(\frac{n(n+1)}{2}\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=37\), we get \(\frac{37\times38}{2}=37\times19=703\). Therefore, 703 is correct. A value such as 683 can result from an error in multiplication or halving. Exam tip: divide the even factor in \(n(n+1)\) by 2 before multiplying.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=42\), so \(\frac{42\times43}{2}=21\times43=903\). Therefore, 903 is correct. The value 882 can result from incorrectly using \(42\times42/2\); the formula must include \(n+1=43\). Exam tip: Cancel the even number 42 with 2 before multiplying.

(S_{41}=\frac{41\times42}{2}=861), so (n=41). In such questions, check the options in the formula.