Sum of first n natural numbers

(S_{102}=5253), (S_{103}=5356), and (S_{101}=5151), so the value is (5458). Subtract carefully with nearby indices.

The difference is ((n-6)+(n-5)+\cdots+n=7n-21), and (7n-21=1001) gives (n=146). Form the sum of the last seven terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{250}=\frac{250\times251}{2}=125\times251=31375\). Hence, option B is correct. Option A may result from using \(250\) in place of \(251\). Exam tip: when \(n\) is even, divide \(n\) by 2 first for quicker calculation.

Here, \(S_n=1+2+\cdots+n=\frac{n(n+1)}{2}\). Thus, \(S_{180}=\frac{180\times181}{2}=16290\), \(S_{170}=\frac{170\times171}{2}=14535\), and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, \(S_{180}-S_{170}+S_{20}=16290-14535+210=1965\). The option 1975 may result from an error of 10 during subtraction or addition. Exam tip: \(S_{180}-S_{170}\) can also be treated as the sum of the terms from 171 to 180.

The direct answer is option A, 1230. For the first n natural numbers, S_n=n(n+1)/2. First solve n(n+1)/2=4656, or n(n+1)=9312. Since 96×97=9312, n=96. Now S_{n+12}-S_n means the sum of the next 12 natural numbers after 96: 97+98+...+108. There are 12 terms, and their average is (97+108)/2=205/2=102.5. Thus the sum is 12×102.5=1230. Equivalently, S_108-S_96=(108×109-96×97)/2=1230. Option A works. Option B, 1240, option C, 1250, and option D, 1260 are all larger than the correctly calculated sum. A common error is to add 96 itself or to use 12 terms ending at 109. Memory cue: S_{n+k}-S_n contains terms n+1 through n+k.

(S_{244}=29890) and (S_{245}=30135), so it first exceeds (30000) at (n=245). Checking nearby values is the safest method.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers, so \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{48}=\frac{48\times49}{2}=1176\), \(S_{96}=\frac{96\times97}{2}=4656\), and \(S_{32}=\frac{32\times33}{2}=528\). Therefore, \(S_{48}+S_{96}-S_{32}=1176+4656-528=5304\). The value 5284 can result from an arithmetic error in addition or subtraction. Exam tip: calculate each \(S_n\) separately before carrying out the final operation.

The sum of the first 175 natural numbers is \\(S_{175}=\frac{175\times176}{2}=15400\\), while the sum of the first 125 natural numbers is \\(S_{125}=\frac{125\times126}{2}=7875\\). Therefore, the difference is \\(15400-7875=7525\\). This is also the sum of the numbers from 126 to 175; 7475 results from an arithmetic error in subtraction. Exam tip: use \\(S_n=\frac{n(n+1)}{2}\\) for each sum and then subtract.

First (n=154), and (S_m=7503) gives (m=122), so (n-m=32). The difference of consecutive sums gives the index.

Direct answer: option C, \(S_{140}\). By definition, \(S_{132}=1+2+\cdots+132\). Adding \(133+134+\cdots+140\) continues the same sequence without a gap, so the complete sum becomes \(1+2+\cdots+140=S_{140}\). Option A, \(S_{136}\), stops four terms too early and omits 137, 138, 139 and 140. Option B, \(S_{138}\), still omits 139 and 140. Option C includes every term from 1 through 140, so it is correct. Option D, \(S_{142}\), would include two extra terms, 141 and 142, which were not added. No numerical calculation is needed; simply inspect the first sum and the consecutive added terms. Memory cue: when a partial natural-number sum is extended continuously, the last included number becomes the new subscript.

(S_{153}=\frac{153\times154}{2}=11781), so (n=153). For large values, using (2S_n) makes identification easier.

(S_{37}=703), (S_{74}=2775), and (S_{111}=6216), so the total is (9694). Add the three different indices carefully.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=13695\), so \(n(n+1)=27390\). Since \(165\times166=27390\), we get \(n=165\). Therefore, \(n-30=165-30=135\). Choosing 140 would mean \(n=170\), whose sum is not 13695. Exam tip: when a sum is given, first write \(2S=n(n+1)\) and look for two consecutive factors.

The difference is ((n-2)+(n-1)+n+(n+1)+(n+2)+(n+3)=6n+3), and (6n+3=615) gives (n=102). Form the sum of the six middle terms.

The direct answer is option C, n=200. Use S_n=n(n+1)/2. We are given n(n+1)/2=20100, so n(n+1)=40200. The consecutive numbers 200 and 201 have product 200×201=40200. Therefore n=200, and checking gives S_200=200×201/2=20100. Option A, 198, gives 198×199/2=19701, not 20100. Option B, 199, gives 199×200/2=19900, also too small. Option C works exactly. Option D, 201, gives 201×202/2=20301, which is too large. The sum increases with n, so the neighboring options confirm the result. Do not confuse n with n+1: the answer is the first factor in the consecutive product. Memory cue: double 20100 and find consecutive factors 200 and 201.

(S_{178}=15931), so (S_{181}=15931+179+180+181=16471). To move three steps ahead, add the next three terms.

There are \(95-41+1=55\) terms from 41 to 95. Hence, the sum is \(\frac{55}{2}(41+95)=\frac{55}{2}\times136=3740\). Therefore, option A is correct. An answer such as 3720 can result from not counting both endpoints correctly. Exam tip: for consecutive integers, always use \(\text{last}-\text{first}+1\) to count the terms.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_n-S_{n-5}\) leaves only the last five terms: \((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10\). Thus, \(5n-10=490\), so \(5n=500\) and \(n=100\). If \(n=98\), the sum would be \(480\), so it is not correct. Exam tip: \(S_n-S_{n-k}\) contains the \(k\) terms from \(n-k+1\) to \(n\).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_{n+3}-S_n\) leaves only the next three terms: \((n+1)+(n+2)+(n+3)=3n+6\). Thus, \(3n+6=222\), so \(3n=216\) and \(n=72\). If \(n=74\), the difference would be \(228\), not 222. Exam tip: In the difference of two consecutive partial sums, cancel the common terms and add only the extra terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{120}=\frac{120\times121}{2}=7260\) and \(S_{85}=\frac{85\times86}{2}=3655\). Therefore, the difference is \(7260-3655=3605\). An answer such as 3585 usually results from a minor subtraction or multiplication error. Exam tip: this difference can also be found directly as the sum of the integers from \(86\) to \(120\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(S_n=5050\), we get \(\frac{n(n+1)}{2}=5050\), so \(n=100\). Hence, \(S_{n+4}-S_n\) is the sum of the next four terms: \(101+102+103+104=410\). The value 406 is only the sum of the next three terms, \(101+102+103\). Exam tip: write \(S_{n+k}-S_n\) as the sum of terms from \(n+1\) to \(n+k\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{64}=\frac{64\times65}{2}=2080\) and \(S_{32}=\frac{32\times33}{2}=528\). Therefore, \(S_{64}:S_{32}=2080:528=130:33\). The ratio \(65:33\) is incorrect because 2080 and 528 cannot both be divided by 32. Exam tip: always reduce a ratio using the greatest common factor of its two terms.

There are \(175-126+1=50\) terms from 126 to 175. Using the arithmetic progression sum formula, \(S=\frac{n}{2}(a+l)\), we get \(S=\frac{50}{2}(126+175)=25\times301=7525\). Hence, option A is correct. An answer such as 7475 can result from incorrectly counting the terms or mishandling an endpoint. Exam tip: for an inclusive range, always add \(+1\) while finding the number of terms.

(7875=S_{125}), so (S_{118}=\frac{118\cdot119}{2}=7021). The listed options should be checked carefully.

Here, \(S_r\) denotes the sum of the first \(r\) natural numbers. In \(S_{n+2}-S_{n-2}\), the terms from \(1\) to \(n-2\) cancel, leaving \((n-1)+n+(n+1)+(n+2)=4n+2\). Thus, \(4n+2=402\), so \(4n=400\) and \(n=100\). Hence, 100 is correct. For the closest distractor, \(n=99\) would give a difference of \(398\), not 402. Exam tip: when subtracting partial sums, write only the uncancelled consecutive terms.