Sum of first n natural numbers

Direct answer: option C, 2500. Use Sₙ = n(n + 1)/2. The first sum is S₁₀₀ = 100 × 101/2 = 5050. The second is S₅₀ = 50 × 51/2 = 1275. Because the expression has 2S₅₀, calculate 2 × 1275 = 2550. Then subtract: 5050 − 2550 = 2500. Option A, 2400, and option B, 2450, do not follow from the correct partial sums and may arise from arithmetic slips. Option C is correct because the coefficient 2 is included. Option D, 2550, is merely the value of 2S₅₀ before the final subtraction, so selecting it means the last step was omitted. Memory cue: evaluate every sum and every coefficient before carrying out the final addition or subtraction.

The governing concept is the sum of the first n natural numbers. Since row r contains 4r lamps, the numbers of lamps in rows 1 through 35 are 4, 8, 12, and so on, up to 4×35. Factor out 4 from the total: total lamps = 4(1+2+3+…+35). Using the formula S_n = n(n+1)/2, we get S_35 = 35×36/2 = 630. Therefore, the total number of lamps is 4×630 = 2520. Hence option D is correct. The other choices do not result from applying the natural-number sum formula and multiplying by the four lamps-per-row factor.

Direct answer: option C, 77. The basic identity is Sₖ − S₍ₖ₋₁₎ = k, because all terms up to k − 1 cancel and only k remains. Apply it to the first pair: Sₙ − S₍ₙ₋₁₎ = n. Apply it to the second pair with k = n − 4: S₍ₙ₋₄₎ − S₍ₙ₋₅₎ = n − 4. Therefore the whole expression is n + (n − 4) = 2n − 4. Equating this with 150 gives 2n − 4 = 150, so 2n = 154 and n = 77. Option A gives 146, option B gives 148, option C gives 150, and option D gives 152. Thus C is the only valid choice. Memory cue: pair consecutive S terms first; each pair collapses to the newly added endpoint.

Use the triangular-number formula \(S_k=\frac{k(k+1)}2\). To identify \(u\), observe that \(S_{53}=\frac{53\cdot54}{2}=53\cdot27=1431\), so \(u=53\). For the second value, \(S_{81}=\frac{81\cdot82}{2}=81\cdot41=3321\), so \(v=81\). Consequently, \(u+v=53+81=134\), which is option A. The positive indices are appropriate because each \(S_k\) is the sum of the first k natural numbers. The other choices do not follow from the two exact triangular-number representations; direct substitution verifies that 53 and 81 are the required indices.

Direct answer: option C, 106. In Sₙ, the terms run from 1 to n; in S₍ₙ₋₆₎, they run from 1 to n − 6. Subtracting cancels all common terms and leaves the six terms n − 5, n − 4, n − 3, n − 2, n − 1, and n. Their sum is 6n − (5 + 4 + 3 + 2 + 1 + 0) = 6n − 15. Set this equal to 621: 6n − 15 = 621, so 6n = 636 and n = 106. Option A gives 609, option B gives 615, option C gives 621, and option D gives 627. Therefore C is correct. The common mistake is to write the answer as 6n, forgetting the five offsets below n. Memory cue: subtracting S₍ₙ₋ₖ₎ leaves the last k terms, whose offsets must also be included.

The governing idea is to convert a consecutive range into a difference of two initial sums. The required terms from 121 through 160 are obtained as \(S_{160}-S_{120}\), because subtracting the sum through 120 removes all terms before 121. Using \(S_n=\frac{n(n+1)}2\), \(S_{160}=\frac{160\cdot161}{2}=12880\) and \(S_{120}=\frac{120\cdot121}{2}=7260\). Thus the required sum is \(12880-7260=5620\), so option D is correct. An independent check uses 40 terms: their average is \((121+160)/2=140.5\), and \(40\cdot140.5=5620\).

Answer: A, 676. The sum of the first n natural numbers is Sₙ = n(n + 1)/2. We need first to identify n. Since 80 × 81/2 = 40 × 81 = 3240, n = 80. Now Sₙ₊₈ − Sₙ means that the common terms from 1 through n cancel, leaving the next eight numbers: 81, 82, 83, 84, 85, 86, 87, and 88. Their sum can be found by the arithmetic-series formula: number of terms × average of first and last = 8 × (81 + 88)/2 = 8 × 169/2 = 676. Therefore option A is correct. Option B, 656, is too small and would come from an incorrect endpoint or arithmetic error. Option C, 696, and option D, 716, similarly do not equal the sum of exactly these eight consecutive terms. Remember: Sₐ − Sᵦ leaves the terms from b + 1 through a, not b through a.

The number of vehicles in row r is \(5r\). Therefore, for 42 rows, the total is \(5(1+2+\cdots+42)=5S_{42}\). Apply the formula \(S_n=\frac{n(n+1)}2\): \(S_{42}=\frac{42\cdot43}{2}=21\cdot43=903\). Multiplying by the common factor 5 gives \(5\cdot903=4515\). Hence option C is correct. The factor 5 applies to every row, not only to the final row, so it can be taken outside the sum. Options A, B, and D are not equal to five times the sum of the first 42 natural numbers and therefore cannot represent the stated arrangement.

Answer: C, 64. By definition, Sₙ = 1 + 2 + ⋯ + n and Sₙ₋₁ = 1 + 2 + ⋯ + (n − 1). When we subtract them, all terms from 1 to n − 1 cancel, leaving only n. Thus Sₙ − Sₙ₋₁ = n, so the first condition gives n = 126. For the second condition, use Sₘ = m(m + 1)/2. We need m(m + 1)/2 = 1953. Since 62 × 63/2 = 31 × 63 = 1953, m = 62. Therefore n − m = 126 − 62 = 64, so option C is correct. Option A would result from an incorrect value of m or n. Option B may result from using m = 64 rather than calculating it. Option D reflects an addition or subtraction error. A useful memory cue is that the difference between two consecutive partial sums is always the newly added term: Sᵣ − Sᵣ₋₁ = r.

Answer: A, 2455. Since Sₙ = 1 + 2 + ⋯ + n, subtracting S₂₄₀ from S₂₅₀ cancels the terms 1 through 240. The terms left are 241 through 250: ten consecutive numbers. Their average is (241 + 250)/2 = 491/2 = 245.5. Therefore their sum is 10 × 245.5 = 2455. We can verify using the formula Sₙ = n(n + 1)/2: S₂₅₀ − S₂₄₀ = [250 × 251 − 240 × 241]/2 = [62750 − 57840]/2 = 4910/2 = 2455. Thus A is correct. Option B is one possible subtraction error, while C and D result from using an incorrect average, endpoint, or number of terms. The common trap is to count 241 through 250 as nine terms; inclusive counting gives 250 − 241 + 1 = 10 terms.

The governing formula for the sum of the first n natural numbers is S_n = n(n+1)/2. Substituting the given value gives n(n+1)/2 = 9180, so n(n+1) = 18360. We now look for consecutive positive integers whose product is 18360. Since 135 × 136 = 18360, n = 135 satisfies the equation. A direct check gives S_135 = 135 × 136/2 = 135 × 68 = 9180. Because n represents the number of terms, it must be a positive natural number, so this solution is valid and unique. Thus option B is correct. The other choices give different consecutive products and therefore cannot produce the stated sum.

Use the governing identity S_n = n(n+1)/2. First calculate S_72 = 72 × 73/2 = 36 × 73 = 2628. Next, S_36 = 36 × 37/2 = 18 × 37 = 666. Their difference is S_72 − S_36 = 2628 − 666 = 1962. The wording asks for twice this difference, not merely the difference, so the required value is 2 × 1962 = 3924. Therefore option C is correct. A common distractor is 1962, which omits the final multiplication, while the nearby listed values can arise from an incorrect sum formula or a subtraction error. Keeping the order of operations clear gives the unique answer.

The governing concept is the sum of a consecutive range using S_n = n(n+1)/2. The numbers from 61 through 125 can be represented as S_125 − S_60, because subtracting the sum through 60 removes all earlier terms. Now S_125 = 125 × 126/2 = 7875, and S_60 = 60 × 61/2 = 1830. Hence the required sum is 7875 − 1830 = 6045. A second check gives 125 − 61 + 1 = 65 terms, with average (61 + 125)/2 = 93, so 65 × 93 = 6045 as well. Therefore option B is correct. The other options usually result from omitting an endpoint or subtracting the wrong preceding sum.

The number of lamps increases in direct proportion to the row number: row 1 has 6 lamps, row 2 has 12, and row r has 6r. Therefore, the total for 45 rows is 6 times the sum of the first 45 natural numbers. This is an arithmetic progression whose first term is 6, last term is 270, and number of terms is 45.

Using the sum formula, the total is \(6(1+2+\cdots+45)=6\times\frac{45(45+1)}{2}\). Thus it is \(6\times\frac{45\times46}{2}=6\times1035=6210\). Hence option B follows. The incomplete supplied explanation points toward the right method but does not finish the calculation; the answer itself is consistent.

Answer: A, 10830. A sum beginning at 151 can be written as a difference of cumulative sums: 151 + ⋯ + 210 = S₂₁₀ − S₁₅₀. This works because S₂₁₀ contains 1 through 210, and subtracting S₁₅₀ removes 1 through 150. Using Sₙ = n(n + 1)/2, S₂₁₀ = 210 × 211/2 = 105 × 211 = 22155, while S₁₅₀ = 150 × 151/2 = 75 × 151 = 11325. Thus the required sum is 22155 − 11325 = 10830. As a second check, there are 210 − 151 + 1 = 60 terms. Their average is (151 + 210)/2 = 180.5, so the sum is 60 × 180.5 = 10830. Therefore A is correct. The other options arise from an endpoint, term-counting, or subtraction error. Inclusive counting is essential because both 151 and 210 are included.

The governing idea is to add the row counts given by the rule 7r. For rows 1 through 56, the total is 7(1 + 2 + 3 + ... + 56) = 7S_56. Using S_n = n(n+1)/2, we find S_56 = 56 × 57/2 = 28 × 57 = 1596. Therefore the total number of vehicles is 7 × 1596 = 11172. Option A is correct. The factor 7 must be applied after summing the row numbers, or equivalently to every row before adding. Options B, C, and D do not follow from the stated rule; they can arise from an arithmetic error in the triangular-number calculation, an incorrect number of rows, or adding an unsupported extra quantity.

Answer: B, 129. The formula for the sum of the first x natural numbers is Sₓ = x(x + 1)/2. Substitute the given value: x(x + 1)/2 = 8385, so x(x + 1) = 16770. We look for two consecutive integers whose product is 16770. Since 129 × 130 = 16770, x = 129. A direct check gives S₁₂₉ = 129 × 130/2 = 129 × 65 = 8385. Hence option B is correct. Checking nearby options also confirms uniqueness: S₁₂₈ = 128 × 129/2 = 8256, whereas S₁₃₀ = 130 × 131/2 = 8515. Since each new natural number increases the sum, no other index between these values can give 8385. Option A is too small and options C and D are too large. The useful pattern is to convert the triangular-number equation into a product of consecutive integers.

Answer: B, 4395. Since S₃₀₀ = 1 + 2 + ⋯ + 300 and S₂₈₅ = 1 + 2 + ⋯ + 285, subtraction cancels all terms through 285. The remaining terms are 286 + 287 + ⋯ + 300. Their count is 300 − 286 + 1 = 15. The average of the first and last terms is (286 + 300)/2 = 293. Therefore the sum is 15 × 293 = 4395. Using the formula gives the same check: S₃₀₀ − S₂₈₅ = 300 × 301/2 − 285 × 286/2 = 45150 − 40755 = 4395. Thus option B is correct. Option A can result from counting fourteen terms, option C from using an incorrect average, and option D from an arithmetic error. The key fact is that S₃₀₀ − S₂₈₅ begins at 286, not 285, because 285 has already been included in the subtracted sum.

Use the triangular-number formula S_n=n(n+1)/2. We have S_96=96×97/2=48×97=4656 and S_48=48×49/2=24×49=1176. Their difference is 4656−1176=3480. The question asks for four times this difference, so 4×3480=13920. Hence option A is correct. A useful alternative is to observe that S_96−S_48=49+50+...+96. This arithmetic sequence has 48 terms and average (49+96)/2=72.5, giving 48×72.5=3480, followed by multiplication by 4. The other choices result from arithmetic slips or from forgetting the final factor of four.

Substitute the given sum into the standard formula: n(n+1)/2=6670. Multiplying both sides by 2 gives n(n+1)=13340. For n=115, the consecutive product is 115×116=13340, so 115×116/2=6670. Therefore n=115 and option C is correct. Checking the other listed values confirms that they fail: 110×111/2=6105, 112×113/2=6328, and 116×117/2=6786. Because n is a natural-number index and the expression n(n+1)/2 strictly increases as n increases, there cannot be another natural-number value among the options producing 6670. Direct substitution provides a complete verification.

First calculate the required triangular sum using S_n=n(n+1)/2. For n=150, S_150=150×151/2=75×151=11325. The percentage concept is that 20% equals 20/100, or one-fifth. Thus 20% of the sum is 11325×20/100=11325/5=2265. Therefore option C is correct. It is important not to take 20% of 150 itself, because the question asks for 20% of the complete sum. A second check is 11325×0.2=2265. The other choices do not result from the correct triangular sum and percentage operation; they reflect an arithmetic error or an incorrect interpretation of what quantity the percentage applies to.

Apply the formula S_n=n(n+1)/2 to the given value. We need n(n+1)/2=8911, so n(n+1)=17822. The consecutive integers 133 and 134 give 133×134=17822. Therefore S_133=133×134/2=8911, which proves that n=133 and option B is correct. Neighboring values confirm the result: S_132=132×133/2=8778 and S_134=134×135/2=9045. The partial sums of positive natural numbers increase strictly, so one index cannot produce the same sum as another. Thus the factorization into consecutive integers is not merely a trial; it establishes the unique natural-number solution.

Use \(S_n=\frac{n(n+1)}{2}\). So \(\frac{n(n+1)}{2}=25,425\Rightarrow n(n+1)=50,850\). Check \(n=225\): \(225\cdot 226=50,850\), so it works. Therefore \(n=225\).

The governing formula for the sum of the first n natural numbers is Sₙ = n(n + 1)/2. Substitute the given value: n(n + 1)/2 = 7503, so n(n + 1) = 15006. Testing the nearby options is efficient: for n = 122, 122 × 123 = 15006, and division by 2 gives 7503. Hence n = 122, so option C is correct. The neighboring values give different sums: n = 121 gives 121 × 122 / 2 = 7381, while n = 123 gives 123 × 124 / 2 = 7626. Since n represents a natural-number count, the positive value is selected; a negative quadratic root would not describe the number of terms.

This is a simultaneous-equation problem involving two consecutive partial sums. Add the equations so that Sₙ₋₁ is eliminated: (Sₙ + Sₙ₋₁) + (Sₙ − Sₙ₋₁) = 5625 + 75. The middle terms cancel, giving 2Sₙ = 5700. Dividing both sides by 2 gives Sₙ = 2850, so option B is correct. Substitution verifies the result: if Sₙ = 2850, then Sₙ₋₁ = 5625 − 2850 = 2775, and 2850 − 2775 = 75. Option A is the corresponding previous partial sum, which is a plausible result of stopping after elimination. Options C and D do not satisfy both given equations, so they are not valid values of Sₙ.