Sum of first n natural numbers

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 \(\frac{n(n+1)}{2}\).

(S_{20}=\frac{20\times21}{2}=210), so (n=20). For a larger sum, checking through options is simple.

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.

\(S_{31}=\frac{31\times32}{2}=496\). If the sequence starts from (1), the last number is (n).

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.