Sum of first n natural numbers

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 the correct answer. Although 656 is close, it is not obtained from the required formula. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of consecutive natural numbers from 1 to \(n\).

The governing concept is the sum of the first n natural numbers: Sₙ = n(n + 1)/2. The natural numbers here are 1, 2, 3, …, 15, so n = 15. Substitution gives S₁₅ = 15(15 + 1)/2 = 15 × 16/2. Dividing 16 by 2 gives 8, and multiplying 15 by 8 gives 120. Therefore option C is correct. One may also pair the terms: (1 + 15), (2 + 14), and so on; each pair totals 16, and the formula efficiently accounts for all terms. Option A or B is too small, while option D is too large, usually because of an incorrect multiplication or an incorrect count of terms.

This is the sum of the first (20) natural numbers, so (S_{20}=210). The dots (\cdots) show that the sequence continues.

The first 7 natural numbers are 1, 2, 3, 4, 5, 6 and 7. Their sum is \(1+2+3+4+5+6+7=28\). Therefore, 28 is correct. The value 21 is the sum of the first 6 natural numbers, so it is a close but incorrect option. Exam tip: the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

(S_9=\frac{9\times10}{2}=45), so (n=9). In such questions, checking options in the formula is fast.

The governing concept is the sum of the first n natural numbers, represented by the triangular-number formula S_n = n(n + 1)/2. The first 12 natural numbers are 1, 2, 3, ..., 12, so n = 12. Substitution gives S_12 = 12 × 13 / 2. Dividing 12 by 2 first gives 6, and then 6 × 13 = 78. Therefore, option C is correct. Option A is too small, option B is not produced by the formula, and option D is too large. The phrase “first 12” means that every integer from 1 through 12 is included; it does not mean 12 × 12 or a list beginning with zero.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=18\) gives \(\frac{18\times19}{2}=171\), so option A is correct. \(190\) is the sum from \(1\) to \(19\), not from \(1\) to \(18\). Exam tip: in \(n(n+1)/2\), use \(n\) as the number of terms given in the question.

The governing concept is the formula for the sum of the first n natural numbers: S_n = n(n + 1)/2. The integers from 1 to 25 are precisely the first 25 natural numbers, so n = 25. Hence S_25 = 25 × 26 / 2 = 25 × 13 = 325. Thus option B is correct. The answer can also be verified by pairing terms: 1 + 25 = 26, 2 + 24 = 26, and so forth. There are 12 complete pairs and the middle number 13 remains, so the total is 12 × 26 + 13 = 325. The other options result from an incorrect multiplication, an incorrect number of terms, or an inaccurate pairing.

The first 5 natural numbers are 1, 2, 3, 4, and 5. Their sum is \(1+2+3+4+5=15\), so option C is correct. The value 10 is the sum of only the first 4 natural numbers. 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 \(\frac{n(n+1)}{2}\). So, \(\frac{n(n+1)}{2}=66\), giving \(n(n+1)=132\). Since \(11\times12=132\), \(n=11\). For \(n=12\), the sum is \(\frac{12\times13}{2}=78\), not 66. Exam tip: look for two consecutive numbers whose product equals \(2S\).

Given \(S_n=\frac{n(n+1)}{2}\). Substituting \(n=6\), \(S_6=\frac{6(6+1)}{2}=\frac{6\times7}{2}=21\). Therefore, the correct answer is 21. Getting 18 usually results from an error in multiplication or division by 2. Exam tip: first find \(n+1\), then multiply and divide by 2.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=30\) gives \(\frac{30\times31}{2}=15\times31=465\). A value such as \(435\) can result from incorrectly using \(30\times29/2\). Exam tip: for the sum of natural numbers, always use \(n+1\) in the formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=14\), so \(\frac{14\times15}{2}=105\). Therefore, the correct answer is 105. The option 110 may result from an incorrect calculation of \(14\times15\div2\). Exam tip: use the last number as \(n\) in the formula.

The first 9 natural numbers are \(1,2,3,\ldots,9\). The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=9\) gives \(\frac{9\times10}{2}=45\). The close distractor 36 is the sum of the first 8 natural numbers. Exam tip: remember that the formula uses both \(n\) and \(n+1\).

The sum of the first (n) natural numbers is (S_n=\frac{n(n+1)}{2}). This formula is the key formula of this topic.

For \(S_n=1+2+\cdots+n\), the correct sum formula is \(\frac{n(n+1)}{2}\). The expression \(\frac{n(n-1)}{2}\) corresponds to the sum up to \(n-1\). Exam tip: check that the factor is \(n+1\), not \(n-1\).

The total is the sum of the dots from the first to the 11th row: \(1+2+3+\cdots+11\). The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(\frac{11\times12}{2}=66\), so 66 is correct. Note that 55 is the sum of only the first 10 natural numbers. Exam tip: For this row pattern, use \(\frac{n(n+1)}{2}\).

The daily readings form the natural-number sequence 1, 2, 3, ..., 16. Therefore the total is the sum of the first 16 natural numbers, not the product of the number of days and a constant daily amount. Use \(S_n=\frac{n(n+1)}{2}\). With n=16, \(S_{16}=\frac{16\times17}{2}=8\times17=136\) pages.

Thus option C is correct. Another way to check is to pair the first and last readings: 1+16=17, 2+15=17, and so on. There are 8 such pairs, giving \(8\times17=136\). The sequence increases by one page each day, so answers such as 126 or 144 do not match the required arithmetic-series total. The supplied answer and explanation are accurate.

The first 4 natural numbers are 1, 2, 3, and 4. Therefore, their sum is 1 + 2 + 3 + 4 = 10. The value 9 is only the sum of 2 + 3 + 4, so it is not correct. Exam tip: The sum of the first n natural numbers is n(n+1)/2.

The governing relation is S_n = n(n + 1)/2. Since the given sum is 91, set n(n + 1)/2 = 91 and multiply by 2 to obtain n(n + 1) = 182. Checking the choices, n = 13 gives 13 × 14 = 182, so S_13 = 182/2 = 91. Therefore option B is correct. The neighboring sums confirm the result: S_12 = 12 × 13/2 = 78, while S_14 = 14 × 15/2 = 105. Because the sum strictly increases as n increases through positive integers, no other listed value can give 91. The number 14 appears as the second factor when n is 13; it is not itself the required value of n.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here \(n=19\), so the sum is \(\frac{19\times20}{2}=190\). Hence, option C is correct. The value 180 may result from an incorrect multiplication of 19 and 20. Exam tip: In the formula, substitute the last natural number for \(n\).

The first (2) natural numbers are (1) and (2), so the sum is (3). Listing terms is safe in basic questions.

The number of bricks in the layers runs from 1 to 10, so the total is \(1+2+3+\cdots+10\). The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{10\times11}{2}=55\), so 55 is correct. The value 45 is the sum from 1 to 9, making it a close but incorrect option. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\) directly.

Direct answer: the sum is 253, so option A is correct. The first 22 natural numbers are 1, 2, 3, ..., 22. The standard formula for the sum of the first n natural numbers is Sₙ = n(n + 1)/2. Here n = 22, so S₂₂ = 22 × 23/2. Divide 22 by 2 first: 11 × 23 = 253. Another useful check is to pair the terms: 1 + 22 = 23, 2 + 21 = 23, and so on. There are 11 such pairs, giving 11 × 23 = 253. A is correct because it equals the formula. B, 242, is too small and can result from an incorrect product or omitted terms. C, 264, may come from using the wrong factor after 22. D, 275, is also larger than the correct total and may result from using an incorrect last term or arithmetic error. Memory cue: for 1 through n, use n(n + 1)/2, not n²/2.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\), so A is correct. \(\frac{n(n-1)}{2}\) gives the sum only up to \(n-1\). Exam tip: test any formula with \(n=1\).