Sum of first n natural numbers

For the sum of the first \\(n\\) natural numbers, \\(S_n=\\frac{n(n+1)}2\\). We are told that \\(S_n=20100\\). Solving \\(\\frac{n(n+1)}2=20100\\) gives \\(n(n+1)=40200\\), and \\(200\\times201=40200\\), so \\(n=200\\). Therefore \\(S_n-S_{n-8}\\) is the sum of the last eight numbers through 200.

Those numbers are 193, 194, 195, 196, 197, 198, 199, and 200. Their sum can be found by pairing the first and last terms: there are 8 terms and each pair totals 393, giving \\(8\\times\\frac{193+200}{2}=8\\times196.5=1572\\). Hence option A is correct.

(1770=S_{59}), so (3n=59), which does not give an integer (n). The correct conclusion is no integer (n).

There are
\(150-50+1=101\) terms from 50 to 150, including both endpoints. Their average is
\(\frac{50+150}{2}=100\). Hence, the sum is
\(101\times100=10100\). The option 10050 results from an error in counting the inclusive terms or finding the average. Exam tip: for an inclusive range from \(a\) to \(b\), use \(b-a+1\) for the number of terms.

Here, \(S_n-S_{n-3}\) represents the sum of the three consecutive natural numbers \(n\), \(n-1\), and \(n-2\). Thus, \(S_n-S_{n-3}=n+(n-1)+(n-2)=3n-3\). From \(3n-3=348\), we get \(n=117\). Therefore, \(S_{117}=\frac{117\times118}{2}=6903\). Hence, option C is correct. The value 6786 may appear close to \(S_{116}\), but the required value is for \(n=117\). Exam tip: \(S_n-S_{n-k}\) is the sum of the last \(k\) terms.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{96}=\frac{96\times97}{2}=4656\) and \(S_{72}=\frac{72\times73}{2}=2628\). Dividing \(4656:2628\) by 24 gives \(194:109\). The ratio \(96:72\) compares only the numbers of terms, not the sums. Exam tip: when comparing such sums, include both factors \(n\) and \(n+1\).

(15400=S_{175}), so (S_{165}=\frac{165\cdot166}{2}=13695). The listed options do not contain the correct value.

Adding the two equations gives (2S_n=10199), so (S_n=5099.5). Such values also help check the consistency of the question.

To add consecutive natural numbers from 201 to 250, use the sum formula for an initial segment and subtract the numbers up to 200. The formula is \\(S_n=\frac{n(n+1)}{2}\\). Subtracting \\(S_{200}\\) removes 1 through 200 and leaves exactly 201 through 250, so no endpoint is missed.

Calculate \\(S_{250}=\frac{250\times251}{2}=31375\\), and \\(S_{200}=\frac{200\times201}{2}=20100\\). Therefore, the required sum is \\(S_{250}-S_{200}=31375-20100=11275\\). A quick check uses 50 terms with average \\(\frac{201+250}{2}=225.5\\), giving \\(50\times225.5=11275\\). Hence option A is correct.

Here, \(S_k\) denotes the sum of the first \(k\) natural numbers. Therefore, \(S_{n+1}-S_{n-4}\) leaves the five terms from \((n-3)\) to \((n+1)\): \[(n-3)+(n-2)+(n-1)+n+(n+1)=5n-5.\] Thus, \(5n-5=595\), so \(5n=600\) and \(n=120\). If \(n=119\), the sum would be \(590\), not \(595\). Exam tip: In \(S_a-S_b\), write the terms from \(b+1\) to \(a\).

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 \(S_{90}=\frac{90\times91}{2}=4095\). Hence, the difference is \(16290-4095=12195\). It is also the sum of the integers from 91 to 180; a value such as 12295 does not result from the formula or subtraction. Exam tip: write such a difference as \(S_{180}-S_{90}\).

Apply S_n=n(n+1)/2 to the given sum. We need n(n+1)/2=8911, or n(n+1)=17822. The consecutive numbers 133 and 134 have product 133×134=17822. Therefore S_133=133×134/2=8911, so n=133 and option B is correct. Checking adjacent choices confirms the result: S_132=132×133/2=8778, while S_134=134×135/2=9045. Since the sequence of partial sums is strictly increasing, no second positive natural-number index can give 8911.

First determine n from n(n+1)/2=7140. Since 119×120/2=7140, n=119. The difference S_(n+5)−S_n consists of the next five natural numbers: 120+121+122+123+124. Their sum is (120+124)×5/2=244×2.5=610. Thus option D is correct. The originally marked option C, 605, is not consistent with the calculation. The difference does not include 119 and includes exactly five terms after n, which is the key indexing point.

The natural numbers 1, 2, 3, …, n form an arithmetic progression. Thus, \(S_n=\frac{n}{2}[2+(n-1)] = \frac{n(n+1)}{2}\). Option B gives the sum up to \(n-1\). Exam tip: check that the final term is n before applying the formula.

(S_{200}-S_{100}=20100-5050=15050), so none of the listed options matches (15150). Always check the difference.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{78}=\frac{78\times79}{2}=3081\), \(S_{52}=\frac{52\times53}{2}=1378\), and \(S_{26}=\frac{26\times27}{2}=351\). Hence, \(S_{78}-S_{52}+S_{26}=3081-1378+351=2054\). Exam tip: evaluate each \(S_n\) separately, then apply the signs in the expression carefully.

(1225=S_{49}), so (2n=98) and (S_{98}=\frac{98\cdot99}{2}=4851). The correct option should be (B).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Thus, \(S_{n+4}-S_n\) leaves only the next four terms: \((n+1)+(n+2)+(n+3)+(n+4)=4n+10\). Therefore, \(4n+10=490\), so \(4n=480\) and \(n=120\). For the closest distractor, \(n=121\) would give a difference of \(494\), not 490. Exam tip: Write \(S_{n+k}-S_n\) as the sum of terms from \(n+1\) to \(n+k\).

Let the sum of the first n natural numbers be \(S_n=\frac{n(n+1)}{2}\). Therefore, the required result is \(S_{175}-S_{125}=\frac{175\times176}{2}-\frac{125\times126}{2}=15400-7875=7525\). It is also the direct sum of the numbers from 126 to 175. An option such as 7475 results from an error in evaluating a sum or subtraction. Exam tip: the difference of two sums with the same starting term equals the sum of the remaining consecutive terms.

(S_{120}:S_{80}=7260:3240=121:54), so the listed options do not contain the correct simplified ratio. Simplify fully.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=11325\), we get \(n=150\). Therefore, \(S_n-S_{n-6}\) is the sum of the last 6 numbers, \(145+146+147+148+149+150=885\). Option 875 is not the correct sum of these six terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last r terms up to n.