Sum of first n natural numbers

The governing concept is the sum of the first n terms of an arithmetic progression. Use S_n = n/2 [2a + (n − 1)d]. With a = 10, d = 2, and n = 6, S6 = 6/2 [2(10) + (6 − 1)(2)] = 3[20 + 10] = 3 × 30 = 90. Thus option C is correct. Direct listing gives the same result: the six terms are 10, 12, 14, 16, 18, and 20, and their sum is 90. Option A may come from omitting or miscalculating terms, option B can result from an incorrect formula substitution, and option D is too large for the listed total. Both the formula and direct addition confirm the answer.

The governing concept is the sum of the first n terms of an arithmetic progression. Use S_n = n/2[2a + (n − 1)d]. Here a = 7, d = 3, and n = 6, so S_6 = 6/2[2(7) + (6 − 1)3] = 3[14 + 15] = 3 × 29 = 87. Thus option C is correct. A direct check lists the terms as 7, 10, 13, 16, 19, and 22. Pairing the first with the last, the second with the fifth, and the third with the fourth gives 29 for every pair; three pairs therefore total 3 × 29 = 87. The other choices can arise from miscounting terms, using an incorrect difference, or making an addition error.

The governing concept is the sum of a finite arithmetic progression. The first term is a = 15 and the common difference is d = 21 − 15 = 6. The first six terms are therefore 15, 21, 27, 33, 39, and 45. Direct addition gives 15 + 21 + 27 + 33 + 39 + 45 = 180. Using the standard formula confirms it: S₆ = 6/2[2(15) + (6 − 1)6] = 3[30 + 30] = 180. Another check is the average of the first and last terms, (15 + 45)/2 = 30, multiplied by six terms: 6 × 30 = 180. Hence option C is correct. The other options reflect an omitted term or an addition error.

The direct answer is option A, \(1176\). The first 48 natural numbers are \(1,2,\ldots,48\). Their sum is \(S_n=\frac{n(n+1)}2\). Substituting \(n=48\), \(S_{48}=\frac{48\times49}{2}=24\times49=1176\). Option A is correct. Option B, 1128, is too small and does not follow from the formula. Option C, 1224, is too large and usually comes from incorrect multiplication. Option D, 1152, may result from forgetting the \(+1\) or using an incomplete calculation. Natural numbers here begin at 1, so zero is not included. Exam cue: for the first \(n\) natural numbers, use \(n(n+1)/2\), and divide the even factor first.

(S_{28}=\frac{28\times29}{2}=406), so (n=28). You can quickly check by substituting the options in the formula.

The numbers of questions solved each day are 1, 2, 3, ..., 24. Therefore, the required total is the sum of the first 24 natural numbers: \(S=\frac{24\times25}{2}=300\). Hence, 300 is correct. Values such as 288 or 324 can result from an incorrect calculation of the sum formula. 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 \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{18}=\frac{18\times19}{2}=171\) and \(S_{12}=\frac{12\times13}{2}=78\). Therefore, \(S_{18}+S_{12}=171+78=249\). Option 239 is incorrect because it is not the sum of these two correct values. Exam tip: In such questions, first use \(S_n=\frac{n(n+1)}{2}\) to find each required sum separately.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{25}=\frac{25\times26}{2}=325\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, \(S_{25}-S_{10}=325-55=270\). This is also the sum of the natural numbers from 11 to 25. Getting 260 usually results from an error while calculating one of the two sums. Exam tip: the difference of two consecutive-range sums directly gives the sum of the terms between them.

(S_{34}=\frac{34\times35}{2}=595), so (n=34). Match the given sum with nearby triangular numbers.

Total chairs are (S_{32}=\frac{32\times33}{2}=528). A pattern increasing by row number forms (S_n).

The sum of the first 40 natural numbers is \\(S=\frac{40\times41}{2}=820\\). Half of this sum is \\(\frac{820}{2}=410\\), so option B is correct. Getting 420 would result from an error while finding the sum or taking half of it. Exam tip: remember that the sum of the first n natural numbers is \\(\frac{n(n+1)}{2}\\).

The direct answer is option A, \(215\). The displayed sequence increases by 1 and ends at 45. Therefore its last five terms are 41, 42, 43, 44, and 45. Add them step by step: \(41+42=83\), \(43+44=87\), and \(83+87+45=215\). Option A is correct. Option B, 220, is 5 greater and may come from adding 46 by mistake. Option C, 225, is also too large and does not represent these five terms. Option D, 230, is larger still and may include an incorrect endpoint. We do not need the entire sum from 1 to 45 because only the final five terms are asked. The average method also works: average \((41+45)/2=43\), then \(5\times43=215\). Memory cue: last five ending at 45 begin at \(45-5+1=41\).

To obtain the sum of the first 21 natural numbers, add the next number, 21, to the sum of the first 20 numbers: \(S_{21}=S_{20}+21=210+21=231\). Hence, 231 is correct. The value 230 would result from adding 20, but the next natural number is 21. Exam tip: For consecutive sums, add the next term to the previous sum.

The sum of the first 36 natural numbers is \,\(S_{36}=\frac{36\times37}{2}=666\), and the sum of the first 30 natural numbers is \,\(S_{30}=\frac{30\times31}{2}=465\). Therefore, the difference is \,\(666-465=201\). It is also the sum of the numbers from 31 to 36. Option 191 would be close to a sum ending at 35, but it does not include 36. Exam tip: The difference of two initial sums equals the sum of the terms left between them.

The direct answer is option B, n=39. The sum of the first n natural numbers is S_n=n(n+1)/2. We need n(n+1)/2=780. Multiplying by 2 gives n(n+1)=1560. For n=39, 39×40=1560, so S_39=39×40/2=780. Option A, 38, gives 38×39/2=741, so it is too small. Option B works exactly. Option C, 40, gives 40×41/2=820, so it is too large. Option D, 41, gives 41×42/2=861, also too large. Since the sum increases when n increases, checking nearby values is enough after using the formula. A common mistake is forgetting the division by 2 or using n^2 instead of n(n+1)/2. Memory cue: double the sum and look for two consecutive factors.

Total dots are (S_{27}=\frac{27\times28}{2}=378). Having (k) dots in the (k)th row shows the sum of natural numbers.

The sum of the first 16 natural numbers is \(S_{16}=\frac{16\times17}{2}=136\). Therefore, \(S_{16}+16=136+16=152\), so option C is correct. The value 136 is only \(S_{16}\); it does not include the additional 16. Exam tip: first use \(S_n=\frac{n(n+1)}{2}\) correctly, then add any extra term separately.

The sum of the first 50 natural numbers is
\(S=\frac{50\times51}{2}=1275\). Therefore, after subtracting 1000, we get \(1275-1000=275\). Hence, option D is correct. A value such as 255 results from an error in finding the sum or carrying out the subtraction. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first k natural numbers is \(S_k=\frac{k(k+1)}{2}\). From \(\frac{m(m+1)}{2}=210\), we get \(m=20\); and from \(\frac{n(n+1)}{2}=325\), we get \(n=25\). Therefore, \(n-m=25-20=5\). Option 4 would result from incorrectly finding the difference between the indices. Exam tip: match a given sum with nearby triangular numbers to identify the index quickly.

The numbers from 11 to 30 form a consecutive arithmetic sequence. A convenient way to add them is to use the sum of the first 30 natural numbers and remove the sum of the first 10 numbers. The formula is \\(S_n=\\frac{n(n+1)}2\\). Thus \\(S_{30}=\\frac{30\\times31}{2}=465\\), and \\(S_{10}=\\frac{10\\times11}{2}=55\\).

The required sum is therefore \\(S_{30}-S_{10}=465-55=410\\). This includes precisely the terms 11 through 30, because the first ten terms, 1 through 10, have been removed. Hence option D is correct. A common mistake is to subtract only the endpoints or to count the terms incorrectly; the difference-of-sums method avoids both errors.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=44\), we get \(\frac{44\times45}{2}=22\times45=990\). Therefore, 990 is correct. A value such as 968 may result from incorrectly using \(44\times44/2\); the formula must include \(n+1\). Exam tip: When \(n\) is even, divide it by 2 first for quicker calculation.

\(S_{n+1}=1+2+\cdots+n+(n+1)\), whereas \(S_n=1+2+\cdots+n\). Thus, the extra term is \(n+1\). Option B incorrectly uses \(n\) as the added term. Exam tip: identify the newly included term in successive sums.

The sum of the first 60 natural numbers includes the last number, 60. Therefore, subtract 60 to get the sum of the first 59 numbers: \(S_{59}=S_{60}-60=1830-60=1770\). Hence, 1770 is correct. Getting 1780 would mean subtracting 50, which is not the last term here. Exam tip: when reducing a consecutive natural-number sum by one term, subtract the omitted last term.

The direct answer is option C, 225. Let S_n denote the sum of the first n natural numbers: S_n=n(n+1)/2. First, S_14=14·15/2=7·15=105. Next, S_15=15·16/2=15·8=120. Adding them gives S_14+S_15=105+120=225. Option A, 210, is not the sum of these two values; it may result from an arithmetic or formula error. Option B, 215, is also too small and has no correct calculation behind it. Option C matches the complete addition. Option D, 235, is too large and can come from adding incorrectly. Do not confuse S_14+S_15 with S_29 or with the sum of only one sequence. The safest method is to calculate both sums separately and then add. Memory cue: use n(n+1)/2 twice, not once.

(S_{42}=\frac{42\times43}{2}=903), so (n=42). Use (2S_n) to see the product of two consecutive numbers.