Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=43\), we get \(\frac{43\times44}{2}=43\times22=946\). Hence, option B is correct. A value such as \(903\) may be associated with the sum up to \(42\), but the question asks for the sum up to \(43\). Exam tip: Divide the even factor \(44\) by 2 before multiplying to calculate faster.

Direct answer: n = 45, so option C is correct. The left side is the sum of the first n natural numbers, so use Sₙ = n(n + 1)/2. We need n(n + 1)/2 = 1035. Multiplying by 2 gives n(n + 1) = 2070. Try n = 45: 45 × 46 = 2070, and therefore 45 × 46/2 = 45 × 23 = 1035. Thus n = 45. An algebraic check gives n² + n − 2070 = 0, which factors as (n − 45)(n + 46) = 0. The roots are 45 and −46; the negative root cannot represent a number of terms. The distractors are close but fail: S₄₃ = 946, S₄₄ = 990, and S₄₆ = 1081. Because the sum strictly increases as n increases, only one natural-number choice can work. Do not confuse the final term n with the total 1035.

There are 10 numbers from 31 to 40, inclusive. They form an arithmetic progression, so the sum is \\(\frac{10}{2}(31+40)=5\times71=355\\). Getting 345 indicates an error in counting the terms or using the last term. Exam tip: When finding the sum over an inclusive range, count both endpoints.

The sum of the first 55 natural numbers is
\(S_{55}=\frac{55\times56}{2}=1540\), and the sum of the first 45 natural numbers is
\(S_{45}=\frac{45\times46}{2}=1035\). Therefore, the difference is
\(1540-1035=505\). Although 500 is close, the exact sum of the numbers from 46 to 55 is 505. Exam tip: For such differences, you can directly add the remaining terms from 46 to 55 instead of finding both full sums.

The direct answer is option B, 1326. Use S_n=n(n+1)/2. We need to identify n from S_n=1225. Since S_49=49·50/2=49·25=1225, n=49. Therefore n+2=51. Now calculate S_51=51·52/2=51·26=1326. Option A, 1275, is S_50, so it stops one term too early. Option B is S_51 and is correct. Option C, 1378, is S_52, so it goes one term too far. Option D, 1431, is S_53, even farther away. Another quick check is S_51=S_49+50+51=1225+101=1326. This is useful because moving from S_n to S_{n+2} means adding the next two numbers, n+1 and n+2. The common mistake is to find n=49 but add only 50 or use S_52. Memory cue: S_{n+2}=S_n+(n+1)+(n+2).

For the sum of the first n natural numbers, \(S_n=1+2+\cdots+n\) and \(S_{n-1}=1+2+\cdots+(n-1)\). Therefore, \(S_n-S_{n-1}=n\). Since the given difference is 42, \(n=42\). If n were 41 or 43, the difference would be 41 or 43, not 42. Exam tip: the difference between consecutive partial sums is always the newly added term.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, the required difference is \(1830-210=1620\). It is also the sum of the natural numbers from 21 to 60. Exam tip: when subtracting two partial sums, check the range of terms left over.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=666\), we get \(n=36\), since \(S_{36}=666\). Therefore, \(S_{n+1}=S_{37}=666+37=703\). The value 693 would result from adding 27, but the next natural number here is 37. Exam tip: use \(S_{n+1}=S_n+(n+1)\) directly.

There are 10 numbers from 41 to 50. Their sum = number of terms × (first term + last term)/2 = \(10 \times (41+50)/2 = 455\). Therefore, 455 is correct. A value such as 445 can result from incorrectly finding the average or the number of terms. Exam tip: for consecutive numbers, number of terms = last term − first term + 1.

\(S_n=1+2+\cdots+(n-1)+n\), whereas \(S_{n-1}=1+2+\cdots+(n-1)\). On subtraction, only \(n\) remains. \(n-1\) is the last term of the previous sum. Exam tip: the difference of consecutive sums gives the newly added term.

There are 10 numbers from 26 to 35. Using the sum formula for an arithmetic progression,

a = 26, d = 1, n = 10


\\(S_n = \frac{n}{2}[2a+(n-1)d]\) = \\(\frac{10}{2}[2(26)+9]\) = \\(5(61) = 305\). Therefore, 305 is the correct answer. A result such as 295 may arise from counting the terms incorrectly or using the wrong last term. Exam tip: For sums of consecutive integers, first determine the number of terms carefully.

Subtracting the sum of the first 60 natural numbers from the sum of the first 70 leaves only the numbers from 61 to 70. Thus, the difference is \(61+62+\cdots+70\). This is an arithmetic progression with 10 terms, so its sum is \(\frac{10}{2}(61+70)=5\times131=655\). Therefore, 655 is correct. 645 is incorrect because the average of the 10 numbers from 61 to 70 is \(65.5\), giving a total of \(10\times65.5=655\). Exam tip: The difference between two “first n” sums equals the sum of the terms left in between.

\(S_n=1+2+\cdots+(n-1)+n\), while \(S_{n-1}=1+2+\cdots+(n-1)\). On subtraction, the common terms cancel and only \(n\) remains. \(2n-1\) is associated with odd-number patterns. Exam tip: write consecutive sums before subtracting.

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), only the last two terms, \((n-1)\) and \(n\), remain. Hence, \(S_n-S_{n-2}=(n-1)+n=2n-1\). Now \(2n-1=79\) gives \(2n=80\), so \(n=40\). If \(n=39\), the value would be \(2(39)-1=77\), not 79. Exam tip: In differences of partial sums, write the terms that remain after cancellation.

The sum of the first 24 natural numbers is \(S_{24}=\frac{24\times25}{2}=300\). Half of this sum is \(300\div2=150\), so 150 is correct. Values such as 140 or 160 do not result from halving the correct total. Exam tip: first use \(\frac{n(n+1)}{2}\) to find the sum, then take the required fraction of it.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Therefore, \(1+2+\cdots+64=\frac{64\times65}{2}=32\times65=2080\). Hence, option C is correct. Option A equals \(\frac{63\times64}{2}=2016\), the sum up to 63, so it omits 64. Exam tip: cancel the factor 2 with the even number first to calculate quickly.

\(S_n=1+2+\cdots+(n-1)+n\), while \(S_{n-1}=1+2+\cdots+(n-1)\). On subtracting, all common terms cancel and only \(n\) remains. \(n-1\) is the previous term, not the difference. Exam tip: compare consecutive sums.

There are 10 consecutive numbers from 51 to 60. Their average is \(\frac{51+60}{2}=55.5\), so the sum is \(10\times55.5=555\). Option 545 is incorrect because it does not result from the correct average of these 10 numbers. Exam tip: for consecutive terms, use \(\text{number of terms}\times\text{average of first and last terms}\).

When the sum of the first 75 natural numbers is subtracted from the sum of the first 80 natural numbers, only the terms from 76 to 80 remain. Thus, the difference is \(76+77+78+79+80=390\). Therefore, option A is correct. Although 395 is close, it is not the sum of these five numbers. Exam tip: In differences of partial sums, cancel the common initial terms and add only the remaining terms.

Direct answer: Sₙ₊₃ = 378, so option B is correct. First find n from Sₙ = n(n + 1)/2 = 300. Since 24 × 25/2 = 300, n = 24. Therefore n + 3 = 27, and the required quantity is S₂₇, not 300 + 3. Now calculate S₂₇ = 27 × 28/2 = 27 × 14 = 378. A second method makes the index shift clear: after the first 24 terms, add the next three terms 25, 26, and 27. Thus S₂₇ = S₂₄ + 25 + 26 + 27 = 300 + 78 = 378. Option A, 351, corresponds to S₂₆; C and D are still larger sums and arise from using an incorrect index or arithmetic. The important habit is to determine n before evaluating Sₙ₊₃. Memory cue: increasing the subscript by 3 means adding the next three natural-number terms, not adding 3 to the sum.

Direct answer: n = 39, so option C is correct. Write the sums explicitly: Sₙ = 1 + 2 + ... + (n − 3) + (n − 2) + (n − 1) + n, while Sₙ₋₃ = 1 + 2 + ... + (n − 3). When the second is subtracted from the first, all common terms cancel, leaving (n − 2) + (n − 1) + n. Add these: n − 2 + n − 1 + n = 3n − 3. The condition becomes 3n − 3 = 114. Add 3: 3n = 117; divide by 3: n = 39. Check: the final three terms are 37, 38, and 39, and their sum is 114. Thus C is correct. For comparison, n = 36 gives 105, n = 38 gives 111, and n = 42 gives 123. Common confusion: subtracting the two sums leaves three terms, not n − 3 or merely the difference of the subscripts.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Pairing terms gives 1+n, 2+(n-1), and so on, each totaling \(n+1\). \(\frac{n(n-1)}{2}\) is the sum only up to \(n-1\). Exam tip: check the last term before selecting a formula.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Multiplying the average, \(\frac{1+n}{2}\), by the number of terms \(n\) gives this result. \(n^2\) is the sum of the first \(n\) odd numbers. In exams, check the division by 2 carefully.

Subtracting the sum of the first 90 natural numbers from the sum of the first 100 leaves only the numbers from 91 to 100. Thus, the sum is \(91+92+\cdots+100=\frac{(91+100)\times10}{2}=955\). Therefore, option B is correct. 945 is a close distractor, but it does not use the correct average, \(95.5\), of the numbers from 91 to 100. Exam tip: In such differences, cancel the common initial terms and write the remaining range directly.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(1176=\frac{48\times49}{2}\), so \(n=48\). Therefore, \(S_{n-1}=S_{47}=\frac{47\times48}{2}=1128\). The value 1176 is \(S_{48}\), not the sum up to the previous term. Exam tip: identify the given sum as half the product of two consecutive numbers first.