Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=95\), so \(1+2+\cdots+95=\frac{95\times96}{2}=95\times48=4560\). A value such as 4655 can result from an incorrect multiplication or addition. Exam tip: whenever the sum from \(1\) to \(n\) is asked, apply \(\frac{n(n+1)}{2}\) directly.

The required terms are 21, 22, ..., 35. One efficient method is to subtract the sum of the first 20 natural numbers from the sum of the first 35 natural numbers. Using \(S_n=n(n+1)/2\), we get \(S_{35}=35\times36/2=630\) and \(S_{20}=20\times21/2=210\). Hence the requested sum is \(S_{35}-S_{20}=630-210=420\). Therefore option D is correct. Equivalently, there are 15 terms, their first and last terms have average \((21+35)/2=28\), and \(15\times28=420\). The other choices result from arithmetic or endpoint-counting errors.

Direct answer: S₃₀ − S₁₈ = 294, so option B is correct. Sₙ denotes 1 + 2 + ... + n, and its formula is Sₙ = n(n + 1)/2. First calculate S₃₀ = 30 × 31/2 = 15 × 31 = 465. Next calculate S₁₈ = 18 × 19/2 = 9 × 19 = 171. Subtract: 465 − 171 = 294. There is a clearer meaning as well: the first 18 terms cancel, leaving 19 + 20 + ... + 30. This contains 30 − 19 + 1 = 12 terms. Its average is (19 + 30)/2 = 24.5, so its sum is 12 × 24.5 = 294. Option A is too small, while C and D result from an arithmetic or endpoint error. Common confusion: S₃₀ − S₁₈ is not the difference between the indices, 12; it is the sum of the terms from 19 through 30.

The governing idea is recognition of consecutive natural numbers. The first expression contains every integer from 1 through 28, and the second expression adds 29, 30, and 31. Together they contain each number from 1 through 31 exactly once. Therefore the total is S₃₁, the sum of the first 31 natural numbers. Using Sₙ = n(n + 1)/2 gives S₃₁ = 31 × 32/2 = 31 × 16 = 496. Hence option A is correct. A separate check gives S₂₈ = 28 × 29/2 = 406 and 29 + 30 + 31 = 90, so 406 + 90 = 496. The other options result from miscounting or arithmetic errors.

The governing idea is the definition of a partial sum: Sₘ = 1 + 2 + 3 + ⋯ + m. The expression already contains every natural number from 1 through 37. Adding the next consecutive terms, 38 and 39, produces 1 + 2 + ⋯ + 37 + 38 + 39, which is exactly the sum of the first 39 natural numbers, S₃₉. Thus option C is correct. It is not S₃₈ because that would stop after adding 38, and it is not S₄₀ because the term 40 has not been included. Option A ignores both added terms. The important step is to track the largest included natural number, not to add the numerical values 38 and 39 to the subscript.

The weekly amounts are 1, 2, 3, …, 49 because the saving starts at 1 rupee and increases by exactly 1 rupee each week. Thus the total is the sum of the first 49 natural numbers. Using Sₙ = n(n + 1)/2, we get S₄₉ = 49 × 50/2 = 49 × 25 = 1225 rupees. Therefore option B is correct. The arithmetic-progression form gives the same result: there are 49 terms, the first is 1 and the last is 49, so total = 49(1 + 49)/2 = 1225. Options A, C, and D result from miscounting weeks or applying the wrong average.

A triangular arrangement with rows containing 1, 2, 3, …, 28 flowers has a total equal to the 28th triangular number. The governing formula is Sₙ = n(n + 1)/2, so S₂₈ = 28 × 29/2 = 14 × 29 = 406. Therefore option B is correct. The word triangular indicates that each new row contributes the next natural number, not that the area of a geometric triangle must be calculated. Option A is smaller than the correct total, while options C and D arise from adding an extra amount or using an incorrect endpoint. Counting the 28 rows and applying the triangular-number formula gives the total directly.

The governing idea is that consecutive terms can be joined into one longer first-n sum. The expression 1 + 2 + ⋯ + 52 already equals S₅₂. Adding 53, 54, and 55 extends the sequence without a gap, so the complete expression is S₅₅. Apply Sₙ = n(n + 1)/2: S₅₅ = 55 × 56/2 = 55 × 28 = 1540. Therefore option A is correct. It would be incorrect to add the indices 52, 53, 54, and 55, because these numbers are terms included in the same sum rather than separate indices to be combined. The other choices are plausible numerical distractors caused by selecting the wrong final term or making a multiplication error.

The notation S_k represents the sum of the first k natural numbers: S_k = 1 + 2 + ... + k. Thus S_75 already contains every term from 1 through 75. The expression adds the five consecutive terms 76, 77, 78, 79, and 80. These are exactly the terms that follow 75, so no number is skipped and no term is repeated. Consequently, S_75 + 76 + 77 + 78 + 79 + 80 = 1 + 2 + ... + 80 = S_80. Therefore option C is correct. S_78 would stop after adding 76, 77, and 78; S_79 would omit 80; and S_81 would require adding one more term, 81, which is not present. The answer follows from tracking the endpoint of the partial sum, not from calculating the numerical total.

For partial sums of natural numbers, subtracting S_190 from S_200 cancels the common terms 1 through 190. Thus S_200 - S_190 = 191 + 192 + ... + 200. There are 10 terms, the first is 191, and the last is 200. Using the arithmetic-progression sum formula, the result is 10/2 × (191 + 200) = 5 × 391 = 1955. Therefore option A is correct. A frequent mistake is to count 191 through 200 as nine terms instead of ten, or to use 190 as the first remaining term; either error can produce one of the nearby distractors.