Sum of first n natural numbers

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 sum is (S_{97}-S_{43}=4753-946=3807). For a middle range, subtract the terms before the start.

(S_{120}-S_{110}=111+112+\cdots+120=1155). The difference of nearby sums gives the sum of terms in between.

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.

(S_{77}=S_{76}+77) and (S_{75}=S_{76}-76), so the value is (1). Use differences of consecutive sums.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{45}=\frac{45\times46}{2}=1035\). Therefore, the required difference is \(4095-1035=3060\). The value 4095 is only the sum of the first 90 natural numbers, not the difference. Exam tip: write \(S_{90}-S_{45}\) before calculating in such questions.

(S_{117}=6903) and (S_{118}=7021), so it first exceeds (7000) at (n=118). In boundary questions, check two nearby values.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{72}=\frac{72\times73}{2}=2628\) and \(S_{24}=\frac{24\times25}{2}=300\). Therefore, \(S_{72}:S_{24}=2628:300=219:25\), so option B is correct. Option \(219:20\) does not result from simplifying the ratio correctly. Exam tip: substitute the value of \(n\) carefully in the sum formula before reducing the ratio.

The difference is ((n-3)+(n-2)+(n-1)+n=4n-6), and (4n-6=282) gives (n=72). Form the sum of the last four terms.

Total tiles are (S_r=2211), and (S_{66}=2211), so there are (66) rows. Such figures form a triangular number.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=2485\), we get \(a=70\), since \(\frac{70\times71}{2}=2485\). Similarly, \(\frac{b(b+1)}{2}=3570\) gives \(b=84\), since \(\frac{84\times85}{2}=3570\). Therefore, \(b-a=84-70=14\). A close distractor such as \(12\) may result from incorrectly identifying the indices. Exam tip: first express each given sum using \(\frac{n(n+1)}{2}\), then determine \(n\).

There are \(112-58+1=55\) terms from 58 to 112. Using the arithmetic progression sum formula, \(S=\frac{n}{2}(a+l)\), we get \(S=\frac{55}{2}(58+112)=\frac{55}{2}\times170=4675\). Hence, 4675 is correct. A value such as 4735 can result from an error in counting terms or finding the average. Exam tip: always include \(+1\) when counting terms from one endpoint to another.

Let S_m be the sum of the first m natural numbers. In S_{m+2}-S_m, only the next two numbers, (m+1) and (m+2), remain. Thus, S_{m+2}-S_m=(m+1)+(m+2)=2m+3. Hence 2m+3=151, so 2m=148 and m=74. If m were 73, the difference would be 149, not 151. Exam tip: In differences of such sums, cancel the common initial terms and add only the extra terms.

The sum of the first 150 natural numbers is \(\frac{150\times151}{2}=11325\), and the sum of the first 125 natural numbers is \(\frac{125\times126}{2}=7875\). Therefore, the difference is \(11325-7875=3450\). It is also the sum of the numbers from 126 to 150. Although 3425 is close, it does not result from the correct calculation. Exam tip: interpret such a difference as the sum of the remaining consecutive terms.

(S_{140}=9870) and (S_{141}=10011), so it first exceeds (10000) at (n=141). In boundary questions, check the previous and next sums.

(S_{35}=630), (S_{55}=1540), and (S_{75}=2850), so the total is (5020). Write all three sums separately and add.

\(S_n=1+2+\cdots+(n-1)+n\), whereas \(S_{n-1}=1+2+\cdots+(n-1)\). On subtraction, all common terms cancel and only \(n\) remains, so A is correct. Option B misses the last term by one. Exam tip: the difference of consecutive partial sums is the newly added term.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{108}=\frac{108\times109}{2}=54\times109=5886\). Hence, 5886 is correct. A value such as 5866 may result from a small multiplication or addition error, but applying the formula gives 5886. Exam tip: When \(n\) is even, divide \(n\) by 2 before multiplying.