Sum of first n natural numbers

(16290=S_{180}), so (S_{172}=\frac{172\cdot173}{2}=14878). Identify (n) first and then reduce it by (8).

In \(S_{n+2}-S_{n-3}\), all terms up to \((n-3)\) cancel. The five remaining terms are \((n-2)+(n-1)+n+(n+1)+(n+2)=5n\). Hence, \(5n=625\), giving \(n=125\). If \(n=126\), the difference would be \(630\), so it is not correct. Exam tip: In differences of partial sums, write and add the uncancelled terms between the two indices.

(S_{200}=20100) and (30%=\frac{3}{10}), so the value is (6030). Convert the percentage into a fraction.

\(S_n\) has the extra final term \(n\) beyond \(S_{n-1}\), so their difference is \(n\). Thus A is correct; B gives the previous term. Exam tip: consecutive sums differ by the newly added term.

The sum of the first 90 natural numbers is \(S_{90}=\frac{90\times91}{2}=4095\). Hence, \(S_n=5775+4095=9870\). Using \(\frac{n(n+1)}{2}=9870\), we get \(n(n+1)=19740=140\times141\); therefore, \(n=140\). For instance, if \(n=135\), the sum is \(\frac{135\times136}{2}=9180\), not the required \(9870\). Exam tip: In such questions, first add the known \(S_k\) to find \(S_n\).

There are \(210-151+1=60\) terms from 151 to 210. Their average is \(\frac{151+210}{2}=180.5\). Hence, the sum is \(60\times180.5=10830\). An option such as 10730 can result from an error in counting the terms or finding the average. Exam tip: when both endpoints are included, always add \(+1\) while counting the terms.

The sum of the first \(r\) natural numbers is \(S_r=\frac{r(r+1)}{2}\). Given \(S_{2n}=4095\), we have \(\frac{2n(2n+1)}{2}=4095\). Since \(4095=\frac{90\times91}{2}=S_{90}\), \(2n=90\), so \(n=45\). If \(n=46\), then the index would be 92, whose sum is not 4095. Exam tip: first identify the index corresponding to the given sum \(S_r\).

Here, \(S_n-S_{n-4}\) represents the sum of the last four natural numbers from \((n-3)\) to \(n\). Thus, \((n-3)+(n-2)+(n-1)+n=4n-6\). Hence, \(4n-6=454\), so \(4n=460\) and \(n=115\). If \(n=114\), the sum is \(450\), so it is the closest option but not correct. Exam tip: In \(S_n-S_{n-k}\), add the last \(k\) terms.

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). Thus,
\(S_{225}=\frac{225\times226}{2}=25425\) and
\(S_{175}=\frac{175\times176}{2}=15400\). Therefore, the difference is
\(25425-15400=10025\). It is also the sum of the integers from 176 to 225. Exam tip: the difference of two such sums equals the sum of the remaining consecutive terms.

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.

(\frac{144\cdot145}{2}=10440), so (n=144). Check the answer by substituting options in the formula.

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 sum is (S_{240}-S_{180}=28920-16290=12630). The subtraction method is quick for large intervals.

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 \(S_n\).

Adding the two equations gives (2S_n=12656), so (S_n=6328). Adding the equations is the fastest way.