Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=63\), we get \(\frac{63\times64}{2}=63\times32=2016\). Hence, option B is correct. An answer such as \(1996\) usually results from an error in multiplication or division. Exam tip: for \(1+2+\cdots+n\), directly use \(\frac{n(n+1)}{2}\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{44}=\frac{44\times45}{2}=990\) and \(S_{30}=\frac{30\times31}{2}=465\). Therefore, \(S_{44}-S_{30}=990-465=525\). This difference is the sum of the natural numbers from 31 to 44. A nearby option such as 535 can result from an arithmetic error in addition or subtraction. Exam tip: interpret \(S_{44}-S_{30}\) directly as the sum from 31 to 44.

The direct answer is option C, n=55. Use S_n=n(n+1)/2. We are told n(n+1)/2=1540, so n(n+1)=3080. The consecutive numbers 55 and 56 have product 55×56=3080. Therefore n=55. Option A, 53, gives 53×54/2=1431, not 1540. Option B, 54, gives 54×55/2=1485, also too small. Option C works because 55×56/2=1540. Option D, 56, gives 56×57/2=1596, which is too large. The index must be a natural number, and the formula uses two consecutive factors n and n+1, not two equal factors. The correct answer is therefore unambiguous. Memory cue: for a triangular-number sum, match 2S with consecutive numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{21}=\frac{21\times22}{2}=231\). Therefore, the required value is \(4\times231=924\). A choice such as \(904\) may result from an error while finding the sum or multiplying it. Exam tip: calculate \(n(n+1)/2\) first, then multiply by the given factor.

Among two consecutive numbers, one is even and the other is odd. Thus \(n(n+1)\) always has a factor of 2, so division by 2 gives an integer sum. Exam tip: use the even-odd pattern for consecutive integers.

The sum of the first 39 natural numbers is \(S_{39}=\frac{39\times40}{2}=780\). Therefore, the required difference is \(780-39=741\). Hence, 741 is the correct option. Subtracting 39 removes the last term, so the result is also the sum of the first 38 natural numbers, \(S_{38}\). Exam tip: First use \(\frac{n(n+1)}{2}\) to find the sum, then perform the required subtraction.

There are \(34-23+1=12\) terms from 23 to 34. This is an arithmetic progression, so its sum is \(\frac{12}{2}(23+34)=6\times57=342\). Therefore, 342 is correct. A value such as 352 can result from an error while counting the terms or finding the average. Exam tip: for consecutive numbers, add the first and last terms and multiply by half the number of terms.

For the sum of the first n natural numbers, \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_n-S_{n-1}=n\), since the difference leaves only the newly added term n. Given \(S_n-S_{n-1}=62\), we get \(n=62\). If n were 61, the difference would be 61, not 62. Exam tip: The difference between two consecutive partial sums is always the newly added term of the sequence.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=75\), we get \(\frac{75\times 76}{2}=75\times 38=2850\). The value \(2775\) is the sum of the first 74 natural numbers, so it is a close but incorrect option. Exam tip: In the formula, remember to use \(n+1\) along with \(n\).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. To obtain \(S_{22}\) from \(S_{18}\), add the next four natural numbers, \(19,20,21\), and \(22\): \(S_{22}=171+19+20+21+22=253\). Hence, the correct answer is 253. One must include every term from 19 through 22, not skip any term. Exam tip: When moving from \(S_m\) to \(S_n\), add all terms from \(m+1\) to \(n\).

If \(S_{45}=1+2+\cdots+45\), then adding \(46\), \(47\), and \(48\) gives \(1+2+\cdots+45+46+47+48=S_{48}\). \(S_{47}\) would include terms only up to 47; adding 48 makes the result \(S_{48}\). Exam tip: when consecutive next natural numbers are added, the subscript of the sum becomes the last number added.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{30}=\frac{30\times31}{2}=465\) and \(S_{15}=\frac{15\times16}{2}=120\). Therefore, \(S_{30}:S_{15}=465:120=31:8\). The ratio \(31:9\) would result from using an incorrect value for the second sum. Exam tip: simplify a ratio at the end by dividing both terms by their greatest common factor.

The difference of two consecutive sums is the new term, so (S_{64}-S_{63}=64). You can answer without finding full sums.

The sum of the first \(t\) natural numbers is \(\frac{t(t+1)}{2}\). Thus, \(\frac{t(t+1)}{2}=231\), so \(t(t+1)=462\). Since \(21\times22=462\), \(t=21\). Therefore, \(t+5=21+5=26\). Option 25 would result from taking \(t=20\), but the sum from 1 to 20 is 210. Exam tip: When a sum is given, first find \(n\) using \(\frac{n(n+1)}{2}\), then evaluate the expression asked.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{68}=\frac{68\times69}{2}=34\times69=2346\). Hence, 2346 is correct. A value such as 2356 can result from an error in multiplication or addition. Exam tip: when \(n\) is even, divide \(n\) by 2 first to calculate quickly.

(S_{13}=91), (S_{16}=136), and (S_{20}=210), so the total is (437). Write the three sums separately and add.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=820\), we get \(a=40\), since \(\frac{40\times41}{2}=820\). Similarly, \(\frac{b(b+1)}{2}=1035\) gives \(b=45\), since \(\frac{45\times46}{2}=1035\). Therefore, \(b-a=45-40=5\). Option 4 may seem close, but it does not equal the actual difference between the indices. Exam tip: equate the given sum to \(\frac{n(n+1)}{2}\) to find the index n.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{80}=\frac{80\times81}{2}=40\times81=3240\). Hence, option B is correct. The other values result from an error in multiplying 80 and 81 or dividing by 2. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of consecutive natural numbers starting from 1.

After the first n numbers, the next natural number is \(n+1\). Hence \(S_{n+1}=S_n+(n+1)\), so the difference is \(n+1\), not n. Exam tip: compare consecutive sums by identifying the newly added term.

Direct answer: option A, 703. The notation Sₙ means 1 + 2 + … + n, so use Sₙ = n(n + 1)/2 separately for both terms. First, S₂₅ = 25 × 26/2 = 25 × 13 = 325. Next, S₂₇ = 27 × 28/2 = 27 × 14 = 378. Adding the two values gives S₂₅ + S₂₇ = 325 + 378 = 703. Option A is therefore correct. Option B, 713, is ten more than the correct result and can arise from a final addition error. Option C, 723, may come from using an incorrect consecutive factor or adding an extra term. Option D, 733, is also inconsistent with the two calculated sums. Do not add the subscripts first: S₂₅ + S₂₇ does not mean S₅₂. Memory cue: calculate each partial sum first, then combine the answers; a subscript identifies the endpoint of one sum.

When the sum of the first 62 natural numbers is subtracted from the sum of the first 67 natural numbers, only the terms from 63 to 67 remain. Thus, \(63+64+65+66+67=325\). Hence, option B is correct. A value of 315 would result from an incorrect addition of these terms. Exam tip: the difference of two consecutive partial sums equals the sum of the intervening terms.

The governing concept is the sum of the first n natural numbers: S_n = 1 + 2 + ... + n = n(n+1)/2. Since S_n = 1830, we have n(n+1)/2 = 1830, so n(n+1) = 3660. The positive integer satisfying this equation is n = 60, because 60 × 61 / 2 = 1830. Therefore, S_{n+10} - S_n = S_70 - S_60. The first 60 terms cancel, leaving only 61 + 62 + ... + 70. This is an arithmetic progression with 10 terms, first term 61, and last term 70. Its sum is 10/2 × (61 + 70) = 5 × 131 = 655. Thus option B is correct. The other choices arise from using an incorrect value of n, counting the new terms incorrectly, or making an addition error.

For the sequence of natural numbers, S_k denotes 1 + 2 + ... + k. The essential property of consecutive partial sums is S_n - S_{n-1} = n, because the only term added when moving from S_{n-1} to S_n is n itself. Hence the first condition gives n = 84. For the second condition, S_m = m(m+1)/2 = 990. Multiplying by 2 gives m(m+1) = 1980. The positive integer solution is m = 44, since 44 × 45 / 2 = 990. Therefore n - m = 84 - 44 = 40. Option C is correct. Values 38, 39, and 41 can result from treating the consecutive-sum difference as a different term, solving the quadratic relation incorrectly, or subtracting the indices inaccurately. Both conditions must be interpreted using the same natural-number partial-sum definition.

Use the governing formula for the sum of the first k natural numbers: S_k = k(k+1)/2. First calculate S_48 = 48 × 49 / 2 = 24 × 49 = 1176. Next calculate S_24 = 24 × 25 / 2 = 12 × 25 = 300. Their difference is S_48 - S_24 = 1176 - 300 = 876. The question asks for three times this difference, so 3 × 876 = 2628. Hence option B is correct. A useful equivalent interpretation is that S_48 - S_24 is the sum of all terms from 25 through 48; this gives the same value. The other choices indicate an error in one partial sum, a failure to multiply by 3, or an arithmetic mistake in the final step.

Direct answer: option C, 72. The sum Sₙ contains terms 1 through n, while S₍ₙ₋₄₎ contains terms 1 through n − 4. On subtraction, the common terms cancel, leaving four terms: (n − 3) + (n − 2) + (n − 1) + n. Combining them gives 4n − 6. Set this equal to the given value: 4n − 6 = 282. Adding 6 gives 4n = 288, and dividing by 4 gives n = 72. Option A, 70, gives 274; option B, 71, gives 278; option C gives 282 exactly; option D, 73, gives 286. Thus only C works. A common confusion is to write the remaining terms as n − 4 through n, which would incorrectly leave five terms. Memory cue: subtracting S₍ₙ₋₄₎ leaves the last four terms of Sₙ.