Sum of first n natural numbers

Correct answer: A. \(285\)

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_{50}-S_{44}\) contains only the terms from 45 to 50: \(45+46+47+48+49+50=285\). Hence, adding \(285\) to \(S_{44}\) gives \(S_{50}\). An option such as \(279\) is incorrect because it is not the complete sum of these six consecutive terms. Exam tip: To find \(S_b-S_a\), add the terms from \(a+1\) to \(b\).

Correct answer: C. (121)

(S_{49}=1225) and (S_{72}=2628), so (u+v=121). First identify the indices of both triangular numbers.

Correct answer: A. (979)

(S_{88}=3916), and (25%), or (\frac{1}{4}), of it is (979). Converting percent to a fraction makes it easier.

Correct answer: C. (1887)

(S_{58}=1711), (S_{59}=1770), and (S_{57}=1653), so the value is (1887). Subtract carefully with nearby indices.

Correct answer: C. (78)

The difference is ((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10), so (5n-10=390) gives (n=80). Form the sum of the last five terms.

Correct answer: B. 11325

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{150}=\frac{150\times151}{2}=75\times151=11325\). Hence, 11325 is correct. A value such as 11225 may result from incorrectly using \(150\times150/2\); the formula must include \(n+1\). Exam tip: write \(\frac{n(n+1)}{2}\) first, then simplify the calculation.

Correct answer: A. 1010

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=5050\), \(S_{90}=4095\), and \(S_{10}=55\). Therefore, \(S_{100}-S_{90}+S_{10}=5050-4095+55=1010\). The value 955 is only \(S_{100}-S_{90}\); \(S_{10}\) must still be added. Exam tip: \(S_a-S_b\) can also be checked as the sum of integers from \(b+1\) to \(a\).

Correct answer: C. \(16\)

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_{3k}=1176\), so \(\frac{3k(3k+1)}{2}=1176\). This gives \(3k=48\), since \(\frac{48\times49}{2}=1176\). Hence, \(k=\frac{48}{3}=16\). If \(k=15\), the index would be \(45\), whose sum is \(1035\), not \(1176\). Exam tip: first find the index \(n\) from the sum, then use \(n=3k\).

Correct answer: B. 2565

These are 30 consecutive natural numbers from 71 to 100. Therefore, their sum is \(\frac{30}{2}(71+100)=15\times171=2565\). Hence, option B is correct. A value such as 2535 can result from an error in counting the number of terms or finding the average. Exam tip: for consecutive numbers, first count the terms using \((\text{last}-\text{first}+1)\).

Correct answer: C. (100)

(S_{99}=4950) and (S_{100}=5050), so it first exceeds (5000) at (n=100). In boundary questions, check nearby values.

Correct answer: B. 1640

The numbers of vehicles in the rows are 2, 4, 6, ..., 80, so the number in each row is twice its row number. Hence the total is \(2(1+2+3+\cdots+40)\). Since \(1+2+\cdots+40=\frac{40\times41}{2}=820\), the total is \(2\times820=1640\) vehicles. The value 1600 may result from using \(40\times40\), but that is not the correct sum of this arithmetic progression. Exam tip: identify the first term, last term and number of terms, then use \(S_n=\frac{n}{2}(a+l)\).

Correct answer: A. \(\left\{\frac{n(n+1)}{2}:n\in\mathbb{N},\ n\geq1\right\}\)

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\), so option A gives the set of triangular numbers. For example, \(n=3\) gives 6. Squares form a different set. In exams, identify the standard sum formula before choosing.

Correct answer: B. 2519

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{33}=\frac{33\times34}{2}=561\), \(S_{66}=\frac{66\times67}{2}=2211\), and \(S_{22}=\frac{22\times23}{2}=253\). Therefore, \(S_{33}+S_{66}-S_{22}=561+2211-253=2519\). An answer such as 2496 can result from calculating one of the sums or the subtraction incorrectly. Exam tip: evaluate each \(S_n\) separately before combining the results.

Correct answer: C. 4020

The sum of the first 120 natural numbers is
\(S_{120}=\frac{120\times121}{2}=7260\), and the sum of the first 80 natural numbers is
\(S_{80}=\frac{80\times81}{2}=3240\). Therefore, the difference is
\(7260-3240=4020\). It is also the sum of the numbers from 81 to 120. A value such as 4060 can result from taking an incorrect starting or ending term. Exam tip: use \(S_n=\frac{n(n+1)}{2}\) for each sum before subtracting.

Correct answer: C. (76)

(S_{76}=\frac{76\times77}{2}=2926), so (n=76). For large values, using (2S_n) makes identification easier.

Correct answer: A. (2740)

(S_{19}=190), (S_{39}=780), and (S_{59}=1770), so the total is (2740). Add the three different indices carefully.

Correct answer: B. 86

The sum of the first n natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4656\), so \(n(n+1)=9312\). Since \(96\times97=9312\), \(n=96\). Therefore, \(n-10=96-10=86\). Choosing 84 would mean \(n=94\), whose sum is not 4656. Exam tip: when a sum is given, first find n using \(n(n+1)/2\), then evaluate the expression asked.

Correct answer: B. (75)

Let \\(S_k\\) denote the sum of the first \\(k\\) natural numbers. The difference \\(S_{n+1}-S_{n-1}\\) contains exactly the terms that are present in the first sum but absent from the second. Those two terms are \\(n\\) and \\(n+1\\). Therefore, \\(S_{n+1}-S_{n-1}=n+(n+1)=2n+1\\). The question gives this difference as 151, so \\(2n+1=151\\). Subtracting 1 gives \\(2n=150\\), and dividing by 2 gives \\(n=75\\). Hence option B is correct.

The same result follows from the formula \\(S_k=k(k+1)/2\\), but cancellation of the common terms is simpler and less error-prone. The first sum ends at \\(n+1\\), while the second ends at \\(n-1\\), so only two consecutive terms remain. Values 74 and 76 would make the difference 149 and 153 respectively, not 151. Thus, the supplied answer is mathematically consistent.

Correct answer: B. (111)

The direct answer is option B, \(111\). For the first \(n\) natural numbers, \(S_n=\frac{n(n+1)}2\). We need \(\frac{n(n+1)}2=6216\), so \(n(n+1)=12432\). Since \(111\times112=12432\), \(n=111\). Verification: \(111\times112/2=111\times56=6216\). Option A, 110, gives \(110\times111/2=6105\). Option B is correct. Option C, 112, gives \(112\times113/2=6328\). Option D, 113, gives \(113\times114/2=6441\). Thus the neighboring choices do not satisfy the given sum. A useful method is to double the sum and look for two consecutive factors. Memory cue: identify \(n\) and \(n+1\) whose product is \(2S_n\).

Correct answer: C. 5050

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{y(y+1)}{2}=4950\), we get \(y=99\). Therefore, \(S_{y+1}=S_{100}=S_{99}+100=4950+100=5050\). Getting 5025 would require adding an incorrect next term. Exam tip: use \(S_{n+1}=S_n+(n+1)\) to solve such questions quickly.

Correct answer: B. 3486

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{83}=\frac{83\times84}{2}=83\times42=3486\). Hence, option B is correct. Option A is close to the sum up to \(82\), while the other values do not follow from the formula. Exam tip: use \(n(n+1)/2\) for the sum of consecutive natural numbers from 1 to \(n\).

Correct answer: C. (85)

(S_{85}=\frac{85\times86}{2}=3655), so (n=85). Match (2S_n) with the product of consecutive numbers.

Correct answer: D. (3807)

The required integers are 44, 45, ..., 97. For a range beginning at 44, it is convenient to subtract the sum of integers through 43 from the sum through 97. The formula for the first n natural numbers is \\(S_n=\frac{n(n+1)}{2}\\). Therefore, \\(S_{97}=\frac{97\cdot98}{2}=4753\\), and \\(S_{43}=\frac{43\cdot44}{2}=946\\). Their difference is \\(4753-946=3807\\).

Hence option D is correct. Subtracting \\(S_{44}\\) would remove 44 itself and give the wrong range, so the upper limit before the starting number must be 43. A quick check also helps: there are 54 terms, and their average is \\(\frac{44+97}{2}=70.5\\); multiplying gives \\(54\times70.5=3807\\).

Correct answer: A. (1155)

The direct answer is option A, \(1155\). By definition, \(S_{120}=1+2+\cdots+120\) and \(S_{110}=1+2+\cdots+110\). Subtracting cancels the common terms 1 through 110, leaving \(111+112+\cdots+120\), ten terms. Pair the first and last: each pair has sum \(111+120=231\); there are five pairs, so the total is \(5\times231=1155\). Option A is correct. Option B, 1165, is 10 too high; option C, 1175, and option D, 1185, likewise do not equal the remaining-term sum. Using formulas also gives \(\frac{120\times121}{2}-\frac{110\times111}{2}=7260-6105=1155\). Memory cue: \(S_m-S_k\) leaves terms \(k+1\) through \(m\).

Correct answer: B. 1711

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=1326\), we get \(n=51\). Hence, \(n+7=58\), and \(S_{58}=\frac{58\times59}{2}=1711\). Option 1671 is not correct because it is not the sum of natural numbers up to 58. Exam tip: first determine \(n\) from the given sum, then apply the formula at the new index.