Sum of first n natural numbers

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 \(b\).

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

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

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

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.

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.

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 \(a\).

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 \(n=3k\).

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 \((\text{last}-\text{first}+1)\).

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

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 \(S_n=\frac{n}{2}(a+l)\).

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.

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.

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.

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

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

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.

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.

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 \(2S_n\).

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.

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 \(n\).

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

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\\).

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 \(m\).

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.