Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=47\), we get \(\frac{47\times48}{2}=47\times24=1128\). Therefore, option B is correct. A value such as \(1104\) can result from using an incorrect number of terms or multiplication. Exam tip: always write both \(n\) and \(n+1\) in the formula.

The governing relationship is Sₙ = n(n + 1)/2 for the first n natural numbers. We seek the option whose product with its successor, divided by 2, equals 561. For n = 33, S₃₃ = 33 × 34/2 = 33 × 17 = 561. Therefore n = 33 and option C is correct. The neighboring choices confirm the uniqueness: S₃₂ = 32 × 33/2 = 528, whereas S₃₄ = 34 × 35/2 = 595. Since successive partial sums differ by the next natural number, the target 561 lies exactly at S₃₃. A common error is to use n²/2 or to forget the factor n + 1; the complete formula must be applied.

The direct answer is option A, 369. The numbers from 12 through 29 include both endpoints. The sum from 1 to n is S_n=n(n+1)/2. The sum from 1 to 29 is S_29=29·30/2=435. To remove the numbers 1 through 11, calculate S_11=11·12/2=66. Thus the required sum is S_29−S_11=435−66=369. Option A is correct. Option B, 379, is 10 more and may come from an addition error. Option C, 389, is 20 more and is also not obtained by the formula. Option D, 399, is 30 more and is too large. A second check uses the arithmetic-series formula: there are 29−12+1=18 terms, and their average is (12+29)/2=20.5; hence the sum is 18·20.5=369. Be careful not to subtract S_12, because 12 must be included. Memory cue: for a to b, use S_b−S_{a−1}.

Total questions are (S_{38}=\frac{38\times39}{2}=741). A daily increase of (1) forms the sum of natural numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{22}=\frac{22\times23}{2}=253\) and \(S_{17}=\frac{17\times18}{2}=153\). Therefore, \(S_{22}+S_{17}=253+153=406\). Although 396 is close, it is not obtained by adding the two correct sums. Exam tip: whenever \(S_n\) appears, use \(\frac{n(n+1)}{2}\) directly.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{41}=\frac{41\times42}{2}=861\) and \(S_{23}=\frac{23\times24}{2}=276\). Therefore, \(S_{41}-S_{23}=861-276=585\). This is also the sum of the natural numbers from 24 to 41. Option 575 is only a nearby value, not the correct difference. Exam tip: You may also directly find the sum from 24 to 41 in such questions.

(S_{43}=\frac{43\times44}{2}=946), so (n=43). Match (2S_n) with the product of two consecutive numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{35}=\frac{35\times36}{2}=630\) and \(S_{19}=\frac{19\times20}{2}=190\). Therefore, \(S_{35}-S_{19}=630-190=440\). It is also the sum of the natural numbers from 20 to 35. Exam tip: \(S_b-S_a\) represents the sum from \(a+1\) to \(b\).

The direct answer is option C, 351 seats. Row 1 has 1 seat, row 2 has 2 seats, and so on; therefore row 26 has 26 seats and the total is 1+2+...+26. This is the sum of the first 26 natural numbers: \(S_{26}=26×27/2=351\). Option A, 325, is not the correct triangular sum. Option B, 338, also does not equal the required total. Option C, 351, exactly equals the calculated total, so it is correct. Option D, 364, is larger than the total. A common mistake is to multiply 26 by 26; the last row gives only the final term, not the sum of all rows. Use \(n(n+1)/2\) whenever the number of seats equals the row number.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(S_{54}=\frac{54\times55}{2}=1485\). Twice this sum is \(2\times1485=2970\). \(2916\) is \(54^2\), so it is not twice the required sum. Exam tip: when the question asks for twice the sum, the 2 cancels in \(\frac{n(n+1)}{2}\), giving \(n(n+1)\).

The sum of the last (6) terms is (35+36+37+38+39+40=225). For last terms, finding full (S_{40}) is not necessary.

To obtain the sum of the first 34 natural numbers, add the numbers after 31, namely 32, 33, and 34, to the given sum. Thus, \(S_{34}=496+32+33+34=595\). Therefore, 595 is correct. Adding only 34 would be incorrect because 32 and 33 are also new terms. Exam tip: When moving from \(S_n\) to \(S_{n+k}\), add every intervening term.

When the sum of the first 55 natural numbers is subtracted from the sum of the first 58 natural numbers, only 56, 57, and 58 remain. Thus, the difference is 56 + 57 + 58 = 171. Option 174 is 3 too large. Exam tip: to find \(S_n-S_m\), add the terms from \(m+1\) to \(n\).

(S_{46}=\frac{46\times47}{2}=1081), so (n=46). Checking with options is useful for larger sums.

Total dots are (S_{42}=\frac{42\times43}{2}=903). Having (r) dots in the (r)th row is the sum of natural numbers.

The sum of the first 28 natural numbers is \(S_{28}=\frac{28\times29}{2}=406\). Therefore, \(S_{28}+28=406+28=434\). Option 435 is \(S_{29}\), because 29—not 28—is added to 406. Exam tip: Use \(S_n=\frac{n(n+1)}{2}\), then add the separately given number carefully.

The sum of the first 65 natural numbers is \(\frac{65\times66}{2}=2145\). Therefore, the required value is \(2145-1500=645\), so option D is correct. A nearby value such as 640 is incorrect because no extra rounding is done after finding the sum and subtracting. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{p(p+1)}{2}=378\), we get \(p=27\), and from \(\frac{q(q+1)}{2}=528\), we get \(q=32\). Therefore, \(q-p=32-27=5\). Option 4 may seem close, but the corresponding indices for 378 and 528 are 27 and 32 respectively. Exam tip: write the sum as \(\frac{n(n+1)}{2}\) and identify the consecutive factors.

There are \(36-18+1=19\) terms from 18 to 36. Using the sum of an arithmetic progression, \(S=\frac{n}{2}(a+l)=\frac{19}{2}(18+36)=19\times27=513\). Hence, option A is correct. A value such as 523 results from using an incorrect number of terms or average. Exam tip: Always add \(+1\) when counting terms from one endpoint to another, inclusive.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{57}=\frac{57\times58}{2}=57\times29=1653\). Hence, 1653 is correct. A value such as 1623 can result from an error while multiplying or halving. Exam tip: remember to use \(n+1\) in the formula.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. From \(\frac{n(n+1)}{2}=666\), we get \(n=36\). Therefore, \(S_{n+2}=S_{38}=S_{36}+37+38=666+75=741\). Option 740 is one less and does not include the correct sum of the next two natural numbers. Exam tip: to find \(S_{n+2}\), add \(n+1\) and \(n+2\) to \(S_n\).

The sum of the first 72 natural numbers includes the last number, 72. Therefore, subtract 72 from 2628 to get the sum of the first 71 natural numbers: \(2628-72=2556\). Option B results from the common mistake of subtracting 71; the number removed is 72. Exam tip: to obtain \(S_{n-1}\) from \(S_n\), subtract \(n\).

Here Sₖ means the sum of the first k natural numbers, so the governing formula is Sₖ = k(k + 1)/2. Calculate both terms separately: S₂₄ = 24 × 25/2 = 300, and S₂₆ = 26 × 27/2 = 351. Their sum is 300 + 351 = 651, so option B is correct. It is important not to interpret S₂₄ + S₂₆ as S₅₀ or as a single sum with index 24 + 26; the notation represents two separate triangular-number sums. Option A omits part of the required addition, while options C and D reflect arithmetic or index errors. Direct substitution into the formula is the safest method.

(S_{48}=\frac{48\times49}{2}=1176), so (n=48). Verify the answer by substituting it back in the formula.

The marks in the first 20 stages are the sum of natural numbers from 1 to 20: \(S_{20}=\frac{20\times21}{2}=210\). The next 5 stages contribute 0 marks, so they do not change the total. Therefore, the total score is 210. A common error is getting 200 by forgetting the \(n+1\) factor in the sum formula. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum from 1 to \(n\).