Sum of first n natural numbers

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.

Since S_n=1+2+...+n, subtracting S_285 from S_300 cancels all terms from 1 through 285. The governing idea is that only the remaining terms 286 through 300 must be added. There are 15 terms, their first and last terms are 286 and 300, and their average is (286+300)/2=293. Therefore the difference is 15×293=4395. Equivalently, S_300−S_285=300×301/2−285×286/2=45150−40755=4395. Thus option B is correct; the other values reflect an incorrect term count or arithmetic error.

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

Use S_n=n(n+1)/2. First, S_96=96×97/2=48×97=4656, and S_48=48×49/2=24×49=1176. Their difference is 4656−1176=3480. The question asks for four times this difference, so 4×3480=13920. Therefore option A is correct. A quicker interpretation is S_96−S_48=49+50+...+96, which also contains 48 terms and has average (49+96)/2=72.5, giving 48×72.5=3480 before multiplying by 4. The other options result from arithmetic slips.

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

Substitute the given value into the sum formula: n(n+1)/2=6670. Multiplying by 2 gives n(n+1)=13340. Among the choices, n=115 gives 115×116=13340, so 115×116/2=6670. Hence option C is correct. The neighboring values do not work: 110×111/2=6105, 112×113/2=6328, and 116×117/2=6786. Since n is a nonnegative counting index and n(n+1)/2 increases with n, no other listed value can satisfy the equation. Direct substitution is therefore sufficient and reliable.

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.

Treat S_n and S_(n−1) as two unknown quantities. Adding the two given equations eliminates S_(n−1): (S_n+S_(n−1))+(S_n−S_(n−1))=4212+65, so 2S_n=4277. Dividing by 2 gives S_n=2138.5, which is option A. Subtracting the equations would give 2S_(n−1)=4147, confirming the algebra. There is a noteworthy consistency issue: if S_n denotes the sum of the first n natural numbers, it should be an integer, so the supplied data cannot correspond to an actual natural-number index n. Nevertheless, the simultaneous equations uniquely determine the listed value.

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.

The sum of the first 150 natural numbers is S_150=150×151/2=75×151=11325. The governing percentage concept is that 20% equals 20/100=1/5. Therefore 20% of the sum is 11325×1/5=2265. Hence option C is correct. It is important to calculate the complete sum before applying the percentage; using 20% of 150 or forgetting the division by 2 would produce unrelated values. The result can also be checked as 11325×20/100=226500/100=2265, confirming the answer.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(\frac{n(n+1)}{2}=12246\). On substituting \(n=156\), we get \(\frac{156\times157}{2}=78\times157=12246\). Hence, 156 is correct. For the closest distractor, \(n=155\) gives \(\frac{155\times156}{2}=12090\), not 12246. Exam tip: when answer choices are given, substituting them in the formula is often the quickest reliable check.

(S_{125}-S_{70}=7875-2485=5390), so none of the listed options matches (5250). Check the difference first.

Subtracting the sum from 1 to 150 from the sum from 1 to 200 leaves only the terms from 151 to 200. Thus, \(S_{200}-S_{150}=\frac{200\times201}{2}-\frac{150\times151}{2}=20100-11325=8775\). Therefore, option B is correct. A value such as 8825 can result from an arithmetic error in the addition or subtraction. Exam tip: recognise such a difference directly as the sum of the remaining consecutive terms, from 151 to 200.