Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=55\), so \(n(n+1)=110\). Since \(10\times11=110\), \(n=10\). For instance, if \(n=9\), the sum is only \(\frac{9\times10}{2}=45\). Exam tip: use \(\frac{n(n+1)}{2}\) directly for sums of consecutive natural numbers.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{16}=\frac{16\times17}{2}=8\times17=136\). Hence, 136 is correct. A value such as 132 results from an incorrect calculation of \(16\times17\) divided by 2. Exam tip: divide the even factor by 2 first to calculate quickly.

The sum of the first \(n\) natural numbers is \(1+2+3+\cdots+n=\frac{n(n+1)}{2}\), so option A is correct. \(\frac{n(n-1)}{2}\) equals the sum of the first \(n-1\) natural numbers, making it a close but incorrect distractor. Exam tip: verify the formula with a small value such as \(n=3\); both the sum and \(\frac{3\times4}{2}\) give \(6\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=13\), we get \(\frac{13\times14}{2}=13\times7=91\). Therefore, 91 is correct. 84 is less than the actual sum of 13 natural numbers. Exam tip: For the sum from 1 to \(n\), directly use \(\frac{n(n+1)}{2}\).

The first 4 natural numbers are 1, 2, 3, and 4. Their sum is \(1+2+3+4=10\), so option C is correct. A value such as 12 may result from counting the numbers incorrectly. Exam tip: the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{22}=\frac{22\times23}{2}=11\times23=253\). The option 242 may result from calculating \(22\times11\), but the correct formula must include \(n+1=23\). Exam tip: Use \(\frac{n(n+1)}{2}\) and divide the even factor by 2 before multiplying.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=4\), \(1+2+3+4=10\), and the formula gives \(\frac{4\times5}{2}=10\). Exam tip: include the \(n+1\) factor.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{24}=\frac{24\times25}{2}=12\times25=300\). Hence, 300 is correct. A value such as \(288\) may result from an incorrect multiplication or addition. Exam tip: Cancel the even factor with 2 first to calculate quickly.

The first three natural numbers are 1, 2, and 3. Therefore, their sum is \(1+2+3=6\). Option 3 is only the third natural number, not the sum. Exam tip: remember that the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=60\), we get \(\frac{60\times61}{2}=30\times61=1830\). A value such as \(1800\) may result from incorrectly using \(60\times60/2\), but the formula must include \(n+1=61\). Exam tip: for consecutive natural numbers starting from 1, use \(\frac{n(n+1)}{2}\).

\(\frac{20\cdot21}{2}=210\), so \(n=20\). Put the options in the formula to find the correct (n).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=17\), we get \(\frac{17\times18}{2}=17\times9=153\), so 153 is correct. The value 170 may result from using \(17\times10\), but that is not the correct sum formula. Exam tip: for the sum of the first \(n\) natural numbers, use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=19\), so \(\frac{19\times20}{2}=19\times10=190\). The values 180, 200, and 210 do not result from this formula. Exam tip: use the last number in the series as \(n\) before applying the formula.

The first 2 natural numbers are 1 and 2. Their sum is \(1+2=3\), so option C is correct. Option 2 is only the second natural number, not the sum. Exam tip: the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

The natural numbers from 1 to n have sum \(\frac{n(n+1)}{2}\). The expression \(\frac{n(n-1)}{2}\) corresponds to the sum up to \(n-1\). Exam tip: substitute n=1; a correct formula must give 1.

Here, \(n\) is the number of consecutive natural numbers being added. For example, in the sum from 1 to 5, \(n=5\). Although 5 is also the largest term, \(n\) denotes the number of terms. Exam tip: identify the term count before using the formula.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{35}=\frac{35\times36}{2}=35\times18=630\). Hence, 630 is correct. Values such as 620 and 640 do not result from this formula. Exam tip: Cancel 2 with the even factor in \(n(n+1)\) before multiplying to calculate quickly.

The sum of the first 10 natural numbers is found using \(S_n = \frac{n(n+1)}{2}\). Here, \(n=10\), so \(S_{10}=\frac{10\times11}{2}=55\). Therefore, 55 is correct. The value 45 is the sum from 1 to 9, so it is a close but incorrect option. Exam tip: For the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\) directly.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=45\), we get \(\frac{45\times46}{2}=45\times23=1035\). Therefore, 1035 is the correct answer. A value such as 1045 may result from an incorrect calculation using the formula. Exam tip: For the sum of consecutive natural numbers, use \(\frac{n(n+1)}{2}\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=325\), so \(n(n+1)=650\). Since \(25\times26=650\), \(n=25\). For \(n=24\), the sum is \(300\), so it is a close but incorrect option. Exam tip: look for consecutive factors of 650 in such questions.