Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{64}=\frac{64\times65}{2}=32\times65=2080\). Hence, 2080 is correct. The value 2048 may result from using \(64\times32\), but that incorrectly omits the \((n+1)=65\) factor. Exam tip: Write \(n(n+1)/2\) and cancel the even factor with 2 first.

(S_{63}=\frac{63\times64}{2}=2016), so (n=63). Match (2S_n) with the product of two consecutive numbers.

The sum is (S_{72}-S_{30}=2628-465=2163). To find a middle range sum, subtract the previous terms.

By definition, S₈₅ contains the terms 1 through 85, while S₈₀ contains the terms 1 through 80. When S₈₀ is subtracted from S₈₅, the common first 80 terms cancel, leaving 81 + 82 + 83 + 84 + 85. This is a sum of five consecutive terms. Using the average method, its value is [(81 + 85)/2] × 5 = 83 × 5 = 415. Therefore option C is correct. The term 80 must not be included in the remainder because it appears in both sums and cancels. The other choices can result from including 80, omitting one remaining term, or making a small addition error.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=465\) gives \(n(n+1)=930\), so \(n=30\). Therefore, \(S_{n+5}=S_{35}=\frac{35\times36}{2}=630\). Options such as \(635\) result from not applying the sum formula correctly for all terms from 31 to 35. Exam tip: first find \(n\) from the given sum, then substitute the new index in the formula.

(S_{51}=S_{50}+51) and (S_{49}=S_{50}-50), so the value is (1). Use differences of consecutive sums.

The sum of the first 40 natural numbers is \(S_{40}=\frac{40\times41}{2}=820\), and the sum of the first 20 natural numbers is \(S_{20}=\frac{20\times21}{2}=210\). Therefore, the required difference is \(820-210=610\), so option C is correct. Getting 600 usually results from an arithmetic error in multiplication or subtraction. Exam tip: In such questions, use \(S_n=\frac{n(n+1)}{2}\) and subtract the two sums.

The direct answer is option C, 71. The expression 1+2+3+⋯+n is the sum of the first n natural numbers, so use S_n=n(n+1)/2. Set n(n+1)/2=2556 and multiply by 2: n(n+1)=5112. For option C, 71·72=5112, so S_71=71·72/2=71·36=2556. Therefore n=71. Option A, 69, gives S_69=69·70/2=2415, which is too small. Option B, 70, gives S_70=70·71/2=2485, also too small. Option C gives exactly 2556. Option D, 72, gives S_72=72·73/2=2628, which is too large. Since the sum increases strictly as n increases, no other listed value can work. A useful pairing check is 1+72=73, 2+71=73, and so on, but the formula is faster. Memory cue: when the final sum is given, solve n(n+1)=2S_n and test nearby options.

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 \(S_{30}=\frac{30\times31}{2}=465\). Hence, \(S_{60}:S_{30}=1830:465=122:31\). Option \(61:15\) may appear close, but it is not the simplified ratio of the two sums. Exam tip: you can also cancel common factors directly in \(\frac{60\times61}{30\times31}\) instead of calculating both sums fully.

(S_n-S_{n-3}=(n-2)+(n-1)+n=3n-3), so (3n-3=144) gives (n=49). Write the difference as the last three terms.

Total blocks are (S_r=1275), and (S_{50}=1275), so there are (50) rows. In such figures, the total objects form a triangular number.

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). From
\(\frac{a(a+1)}{2}=1540\), we get
\(a=55\), and from
\(\frac{b(b+1)}{2}=2080\), we get
\(b=64\). Therefore,
\(b-a=64-55=9\). Although 8 may seem close, the indices corresponding to the two given sums differ by exactly 9. Exam tip: equate the given sum to
\(\frac{n(n+1)}{2}\) and solve for n.

This is an arithmetic sequence with first term 46, last term 90, and \(90-46+1=45\) terms. Therefore, \(S=\frac{45}{2}(46+90)=\frac{45}{2}\times136=3060\). Option 4095 is the sum from \(1\) to \(90\), so it incorrectly includes the terms from \(1\) to \(45\). Exam tip: while counting consecutive integers from one endpoint to another, always add \(1\).

Direct answer: option B, 75. Sₘ is 1 + 2 + … + m, while S₍ₘ₊₁₎ is 1 + 2 + … + m + (m + 1). When Sₘ is subtracted from S₍ₘ₊₁₎, all common terms cancel and only the newly added term m + 1 remains. Thus S₍ₘ₊₁₎ − Sₘ = m + 1. The question states that this difference is 76, so m + 1 = 76 and m = 75. Option A, 74, subtracts too much. Option B is correct because one is subtracted from 76. Option C, 76, results from forgetting the +1 in the new term. Option D, 77, comes from adding rather than subtracting one. Memory cue: the difference between consecutive partial sums equals the newly added term, not the old endpoint.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=\frac{100\times101}{2}=5050\) and \(S_{75}=\frac{75\times76}{2}=2850\). The required difference is \(5050-2850=2200\), so option B is correct. It is also the sum of the integers from 76 to 100. Option 2250 can result from an incorrect value of \(S_{75}\) or a subtraction error. Exam tip: interpret \(S_b-S_a\) as the sum of terms from \(a+1\) to \(b\).

(S_{76}=2926) and (S_{77}=3003), so the sum first exceeds (3000) at (n=77). In boundary questions, check the two nearby sums.

(S_{25}=325), (S_{35}=630), and (S_{45}=1035), so the total is (1990). Write all three sums separately and add.

For the sum of the first n natural numbers, \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3240\), we get \(n=80\). Therefore, \(S_{n-4}=S_{76}\). Now \(S_{76}=S_{80}-(77+78+79+80)=3240-314=2926\). Option 2916 results from subtracting the last four terms incorrectly. Exam tip: To find \(S_{n-k}\), subtract the last k terms from \(S_n\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=96\), \(\frac{96\times97}{2}=48\times97=4656\). Hence, option A is correct. A value such as \(4626\) may result from an error in multiplication or addition. Exam tip: for the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{40}=\frac{40\times41}{2}=820\), \(S_{60}=\frac{60\times61}{2}=1830\), and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_{40}+S_{60}-S_{50}=820+1830-1275=1375\). An answer such as 1385 can result from an error while evaluating one of the sums. Exam tip: calculate each \(S_n\) separately before performing the final addition and subtraction.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_{2p}=820\), so \(\frac{2p(2p+1)}{2}=820\). Since \(\frac{40\times41}{2}=820\), we get \(2p=40\), hence \(p=20\). If \(p=19\), the index would be \(38\), whose sum is not 820. Exam tip: When a sum is given, equate it to \(n(n+1)/2\) and find the index \(n\) first.

Direct answer: option B, 1225. Use Sₙ = n(n + 1)/2. First, S₇₀ = 70 × 71/2 = 35 × 71 = 2485. Next, S₃₅ = 35 × 36/2 = 35 × 18 = 630. The expression contains 2S₃₅, so multiply the entire value by 2: 2S₃₅ = 2 × 630 = 1260. Finally, S₇₀ − 2S₃₅ = 2485 − 1260 = 1225. Option A, 1215, and option C, 1235, can result from small multiplication or subtraction errors. Option B is correct because the coefficient 2 has been applied before subtraction. Option D, 1245, may come from using an incorrect value for S₃₅ or mishandling the coefficient. Memory cue: a number written before S multiplies the whole sum, not just its subscript.

Row r contains 3r lamps, so the total for rows 1 through 30 is 3(1 + 2 + ⋯ + 30), or simply 3S₃₀. Using the first-n sum formula, S₃₀ = 30 × 31/2 = 15 × 31 = 465. Therefore the total number of lamps is 3 × 465 = 1395, so option A is correct. The factor 3 is common to every row and can be taken outside the summation. It should not be added as an extra row or applied only to the last row. The other options may result from calculating S₃₀ incorrectly, mishandling the multiplier, or making an arithmetic error after finding the sum.

(S_{90}=\frac{90\times91}{2}=4095), so (n=90). Matching (2S_n) with a consecutive product is useful.

For sums of the first natural numbers, Sₖ − S₍ₖ₋₁₎ = k. Apply this identity to both pairs: Sₙ − S₍ₙ₋₁₎ = n and S₍ₙ₋₂₎ − S₍ₙ₋₃₎ = n − 2. Thus the complete expression equals n + (n − 2) = 2n − 2. The corrected value is 100, so 2n − 2 = 100, giving 2n = 102 and n = 51. Hence option C is correct. The original value 99 would produce n = 50.5, which is not a natural-number index and makes the original MCQ invalid; it has therefore been corrected to 100. The signs must be grouped as the two consecutive-sum differences shown.