Sum of first n natural numbers

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.

(S_{80}=3240), so (2k+3=80) gives no integer (k). Check the given index condition too.

The sum is (S_{210}-S_{150}=22155-11325=10830). When starting from (151), subtract the sum up to (150).

(S_{96}=4656), so the difference is (97+98+\cdots+108=1230). Find the sum of the next (12) terms separately.

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

Total vehicles are (7S_{56}=7\times1596=11172). In a pattern like (7r), take (7) outside the sum.

(S_{129}=\frac{129\times130}{2}=8385), so (x=129). Use (2S_x) to find the consecutive product.

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.

(S_{132}+133+134+\cdots+140=S_{140}). When consecutive new terms are added, the last term becomes the new index.

(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 difference is (286+287+\cdots+300=4395). Even with large indices, only the terms in between need to be added.

(S_{200}=\frac{200\times201}{2}=20100), so (n=200). Match (2S_n) with a consecutive product.

(S_{96}-S_{48}=4656-1176=3480), and (4\times3480=13920). First find the difference and then multiply.