Sum of first n natural numbers

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_{175}=15400) and (40%=\frac{2}{5}), so the value is (6160). Convert the percentage into a fraction.

\(S_n\) 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.

The sum is (S_{350}-S_{300}=61425-45150=16275). If it starts from (301), subtract the sum up to (300).

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.

(31375=S_{250}), so (S_{238}=\frac{238\cdot239}{2}=28441). Identify (n) first and reduce it by (12).

Using \(S_n=\frac{n(n+1)}{2}\), we get \(8S_n+1=4n(n+1)+1=(2n+1)^2\), which is always an odd perfect square. Exam tip: substitute the sum formula and look for a perfect-square identity.