Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{150}=\frac{150\times151}{2}=11325\) and \(S_{100}=\frac{100\times101}{2}=5050\). Hence, \(S_{150}:S_{100}=11325:5050=453:202\). \(3:2\) is only the ratio of 150 and 100; the factors \(151\) and \(101\) must also be included for the sums. Exam tip: In ratios of such sums, apply \(\frac{n(n+1)}{2}\) first and then cancel common factors.

(15400=S_{175}), so the sum of the last ten terms is (166+167+\cdots+175=1705). Add the last terms directly.

Here, \(S_r\) denotes the sum of the first \(r\) natural numbers. In \(S_{n+1}-S_{n-4}\), the terms from \(1\) to \(n-4\) cancel, leaving \((n-3),(n-2),(n-1),n,(n+1)\). Their sum is \(5n-5\). Hence, \(5n-5=695\Rightarrow 5n=700\Rightarrow n=140\). If \(n=139\), the sum would be \(690\), so it is not correct. Exam tip: When subtracting partial sums, first list the uncancelled terms and count them carefully.

There are \(125-75+1=51\) terms from 75 to 125. Their average is \(\frac{75+125}{2}=100\). Therefore, the sum is \(51\times100=5100\). A value such as 5150 can result from an error in counting the terms or finding the average. Exam tip: Always add \(+1\) when counting terms in an inclusive range.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{120}=\frac{120\times121}{2}=7260\), \(S_{80}=3240\), and \(S_{40}=820\). Therefore, \(S_{120}-S_{80}+S_{40}=7260-3240+820=4840\). The option 4740 can result from an error while adding or subtracting the final \(S_{40}\). In exams, calculate each \(S_n\) separately before substituting.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), only the last two terms, \((n-1)\) and \(n\), remain: \(S_n-S_{n-2}=(n-1)+n=2n-1\). Thus, \(2n-1=257\), giving \(n=129\). Hence, \(S_{129}=\frac{129\times130}{2}=8385\). Option 8256 is \(S_{128}\), so it is close but not correct. Exam tip: For \(S_n-S_{n-2}\), directly add the last two terms.

The required numbers form a consecutive sequence beginning at 181 and ending at 240. The sum of numbers from 181 to 240 can be found efficiently by subtracting the sum through 180 from the sum through 240. Using \\(S_n=\frac{n(n+1)}{2}\\) ensures that both endpoints are handled correctly.

We have \\(S_{240}=\frac{240\times241}{2}=28920\\) and \\(S_{180}=\frac{180\times181}{2}=16290\\). Therefore, the required sum is \\(28920-16290=12630\\). As a second check, there are 60 terms and their average is \\(\frac{181+240}{2}=210.5\\), so the sum is \\(60\times210.5=12630\\). Hence option A is correct.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{128}=\frac{128\times129}{2}=8256\) and \(S_{64}=\frac{64\times65}{2}=2080\). Therefore, \(8256:2080=258:65\), so the correct answer is \(258:65\). \(2:1\) is only an approximate comparison, not the exact simplest ratio. Exam tip: write the sum formula for both terms first, then cancel common factors in the ratio.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=8256\), we get \(n=128\). Therefore, \(S_{n+5}-S_n\) is the sum of the next five numbers: \(129+130+131+132+133=655\). A value such as 645 results from choosing an incorrect starting term. Exam tip: first find \(n\), then list the terms added beyond \(S_n\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=1830\) gives \(n=60\). Therefore, \(3n=180\), and \(S_{3n}=S_{180}=\frac{180\times181}{2}=16290\). A nearby value such as \(12880\) can result from finding \(3n\) incorrectly or applying the sum formula wrongly. Exam tip: first determine \(n\) from \(S_n\), then substitute \(3n\) in the sum formula.

Here, \(S_k\) denotes the sum of the first \(k\) natural numbers. In \(S_{n+3}-S_{n-2}\), the terms from \(1\) to \(n-2\) cancel, leaving \((n-1)+n+(n+1)+(n+2)+(n+3)=5n+5\). Thus, \(5n+5=760\), so \(5n=755\) and \(n=151\). If \(n=152\), the difference would be \(765\), so it is not correct. Exam tip: when subtracting two partial sums, write only the uncancelled consecutive terms.

There are \(150-90+1=61\) terms from 90 to 150. Using the sum of an arithmetic progression, \(\frac{61}{2}(90+150)=\frac{61}{2}\times240=7320\). Hence, option B is correct. Taking only \(150-90=60\) would miss one endpoint. Exam tip: add \(+1\) when both endpoints are included in the range.

\(S_n\) includes 1 through n, whereas \(S_{n-1}\) includes only 1 through \(n-1\). Their difference leaves n: \([n(n+1)-n(n-1)]/2=n\). Exam tip: the difference of consecutive partial sums gives the newly added term.

\(S_n-S_{n-7}\) means that the sum of the first \(n-7\) natural numbers is subtracted from the sum of the first \(n\) natural numbers. The remaining seven terms are \(n-6,n-5,\ldots,n\), whose sum is \(7n-21\). Thus, \(7n-21=1330\), so \(7n=1351\) and \(n=193\). If \(n=194\), the sum would be \(1337\), so it is not correct. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last \(r\) terms.

There are 50 terms from 211 to 260, inclusive. Using the arithmetic progression sum formula, \(S=\frac{n}{2}(\text{first term}+\text{last term})\), we get \(S=\frac{50}{2}(211+260)=25\times471=11775\). Hence, option B is correct. A nearby value such as 11675 may result from counting the number of terms incorrectly. Exam tip: when both endpoints are included, use \(\text{last}-\text{first}+1\) for the number of terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=\frac{100\times101}{2}=5050\) and \(S_{50}=\frac{50\times51}{2}=1275\). Hence, \(S_{100}:S_{50}=5050:1275=202:51\). \(2:1\) may seem close, but it is not correct because the sums also include the \(n+1\) factor. Exam tip: apply the sum formula, cancel common factors, and then reduce the ratio.

Using \(S_n=\frac{n(n+1)}{2}\) and \(S_n=11175\), we get \(n=149\). Hence, \(S_n-S_{n-4}=S_{149}-S_{145}\), which is the sum of the natural numbers from 146 to 149: \(146+147+148+149=590\). Therefore, 590 is correct. A value such as 580 can result from taking one of the last four terms incorrectly. Exam tip: \(S_n-S_{n-k}\) always represents the sum of the last \(k\) terms.

Substituting n=50 gives 2n=100. The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{100}=\frac{100\times101}{2}=5050\) and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_{2n}-S_n=5050-1275=3775\). An option such as 3875 results from an error in subtraction or in applying the sum formula. Exam tip: evaluate the index 2n first, then find the two sums separately.

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), only the last two terms, \((n-1)\) and \(n\), remain. Hence, \(2n-1=199\), giving \(n=100\). Therefore, \(S_{100}=\frac{100\times101}{2}=5050\). Option 4950 is the sum of the first 99 natural numbers, so it is not correct. Exam tip: interpret \(S_n-S_{n-2}\) directly as the sum of the last two terms.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=13041\), so \(n(n+1)=26082\). Since \(161\times162=26082\), \(n=161\). For example, \(n=160\) gives \(\frac{160\times161}{2}=12880\), not 13041. Exam tip: when a sum is given, multiply it by 2 and look for a product of consecutive integers.

Here, \(S_{n+5}-S_{n-5}\) represents the sum of 10 consecutive natural numbers from \((n-4)\) to \((n+5)\). Therefore, \(S_{n+5}-S_{n-5}=10n+5\). So, \(10n+5=1205\), which gives \(10n=1200\) and hence \(n=120\). If \(n=121\), the sum would be \(1215\), not the given value. Exam tip: In such differences, list the first and last remaining terms to verify the number of terms.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{180}=\frac{180\times181}{2}=16290\) and \(S_{120}=\frac{120\times121}{2}=7260\). Therefore, the difference is \(16290-7260=9030\). It is also the sum of the integers from \(121\) to \(180\). Exam tip: The difference of two such sums equals the sum of the terms lying between them.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=2145\), giving \(n=65\), since \(\frac{65\times66}{2}=2145\). Therefore, \(S_{2n}-S_n=S_{130}-S_{65}=\frac{130\times131}{2}-2145=8515-2145=6370\). A value such as 6270 can result from an error in applying the sum formula or in the final subtraction. Exam tip: first determine \(n\) from the given \(S_n\), and then calculate \(S_{2n}\).

There are 50 terms from 401 to 450. Their average is \(\frac{401+450}{2}=425.5\). Therefore, the sum is \(50\times 425.5=21275\). A value such as 21375 can result from a small error in counting terms or finding the average. Exam tip: for an inclusive interval, the number of terms is \(last-first+1\).

For the sum of the first n natural numbers, \(S_n-S_{n-1}=n\), because the new term added to \(S_{n-1}\) is n. Hence, \(n=225\). Therefore, \(S_{n+1}=S_{226}=\frac{226\times227}{2}=25651\). Option A may seem close to \(S_{225}\), but the question asks for the next sum, \(S_{226}\). Exam tip: In such questions, identify \(S_n-S_{n-1}\) directly as n.