Sum of first n natural numbers

Given \(\frac{n(n+1)}{2}=861\), we get \(n(n+1)=1722\). Since \(41\times42=1722\), \(n=41\). For the close distractor \(40\), the sum is \(\frac{40\times41}{2}=820\), not 861. Exam tip: look for two consecutive numbers whose product equals twice the given sum.

To find the sum from 26 to 50, subtract the sum of natural numbers up to 25 from the sum up to 50. Thus, \\(S_{50}-S_{25}=\frac{50\times51}{2}-\frac{25\times26}{2}=1275-325=950\\). Therefore, the correct answer is 950. An option such as 975 can result from using an incorrect number of terms or lower limit. Exam tip: For a range sum, always subtract the sum up to one less than the lower limit.

Here, \(S_n-S_{n-4}\) represents the sum of the four natural numbers from the \((n-3)\)th term to the \(n\)th term: \(S_n-S_{n-4}=(n-3)+(n-2)+(n-1)+n=4n-6\). Hence, \(4n-6=250\), so \(4n=256\) and \(n=64\). Therefore, 64 is the correct option. For instance, if \(n=63\), the sum is \(60+61+62+63=246\), not 250. Exam tip: \(S_n-S_{n-k}\) always gives the sum of the last \(k\) terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{40}=\frac{40\times41}{2}=820\). Therefore, the difference is \(1830-820=1010\). It is also the sum of the numbers from \(41\) to \(60\); \(1000\) does not give the correct difference. Exam tip: the difference of two such sums can be found by adding the remaining consecutive terms directly.

Direct answer: option C, 1431. Since Sₙ is the sum of the first n natural numbers, Sₙ = n(n + 1)/2. The given equation gives n(n + 1)/2 = 1275, so n(n + 1) = 2550. The consecutive numbers 50 and 51 have product 2550, hence n = 50. The required sum is therefore Sₙ₊₃ = S₅₃. Applying the formula again, S₅₃ = 53 × 54/2 = 53 × 27 = 1431. Option A, 1378, does not equal the sum through 53. Option B, 1406, is S₅₂ and reflects increasing the index by only two. Option C is correct because the index must increase from 50 to 53. Option D, 1485, comes from using an incorrect index or arithmetic. Memory cue: first solve for the original index, then apply the requested shift; do not shift the sum value itself.

The sum is \(S_n=\frac{n(n+1)}{2}\). It is odd only when n leaves remainder 1 or 2 on division by 4. Merely knowing that n is odd or even is not enough. Exam tip: check \(n\bmod 4\).

There are \(75-51+1=25\) terms from 51 to 75, inclusive. This is an arithmetic sequence with first term 51 and last term 75. Hence, the sum is \(\frac{25}{2}(51+75)=\frac{25}{2}\times126=1575\). Therefore, option A is correct. An answer such as 1600 can result from an error in counting the terms or finding the average. Exam tip: always add \(+1\) when counting terms from one inclusive endpoint to another.

For Sₙ = 1 + 2 + 3 + ... + n, subtracting S₍ₙ₋₁₎ = 1 + 2 + ... + (n − 1) leaves only the final term n. Hence Sₙ − S₍ₙ₋₁₎ = n. The given difference is 57, so n = 57. Now calculate the required sum: S₅₇ = 57 × 58/2 = 57 × 29 = 1653. Therefore option B is correct. Option A is close but reflects an arithmetic error in the product or division. Options C and D correspond to using a larger index or adding an unrelated term. The important concept is that the difference of two consecutive partial sums identifies the newly included natural number.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{25}=\frac{25\times26}{2}=325\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, the ratio is \(325:55=65:11\). Option 55:13 is incorrect because it is not obtained by simplifying 325:55 correctly. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. From \(\frac{n(n+1)}{2}=2485\), we get \(n=70\), since \(\frac{70\times71}{2}=2485\). Therefore, \(S_n-S_{n-5}=S_{70}-S_{65}\), which is the sum of the last five numbers: \(66,67,68,69,70\). Hence, \(66+67+68+69+70=340\). Choosing 350 would miss the correct first term, 66, among these five terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last \(r\) terms.

The sum of the first 80 natural numbers is \(\frac{80\times81}{2}=3240\), and the sum of the first 50 natural numbers is \(\frac{50\times51}{2}=1275\). Therefore, the required difference is \(3240-1275=1965\). Equivalently, it is the sum of the integers from 51 to 80. Exam tip: use \(n(n+1)/2\) carefully for both sums to avoid a close error such as 1975.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=595\), we get \(n=34\). Therefore, \(S_{n+6}-S_n=S_{40}-S_{34}=35+36+37+38+39+40=225\). Although 230 may seem close, it does not equal the sum of the next six terms. In exams, first find n and then add only the newly included terms.

For \(n=15\), the index \(2n\) becomes 30. Hence, \(S_{2n}=S_{30}=\frac{30(30+1)}{2}=\frac{30\times31}{2}=465\). Option 435 equals \(S_{29}\), so it is not correct. Exam tip: evaluate the index \(2n\) before substituting into the formula.

\(S_{2n}=\frac{2n(2n+1)}{2}=n(2n+1)\), whereas \(2S_n=n(n+1)\). Their difference is \(n^2\), so \(S_{2n}=2S_n+n^2\). Option D ignores this extra \(n^2\) term. Exam tip: substitute \(S_n=\frac{n(n+1)}{2}\) before comparing such expressions.

Adding the two equations gives (2S_n=420), so (S_n=210). When two equations are given, addition is quick.

The sum of the first n natural numbers is \(1+2+\cdots+n=\frac{n(n+1)}{2}\), half the product of consecutive integers n and \(n+1\). It is not always even: for n=2, the sum is 3. Exam tip: check the \(n+1\) factor carefully.

The sum of the first 36 natural numbers is \(S=\frac{36\times 37}{2}=666\). Therefore, \(\frac{1}{3}\) of this sum is \(666\div 3=222\). Although 224 is close, it is not obtained by dividing 666 by 3. Exam tip: first use \(\frac{n(n+1)}{2}\) to find the sum, then apply the given fraction.

(S_{56}-S_{30}=1596-465=1131), so (1110) does not match the listed options. Always check options carefully.

There are 20 numbers from 101 to 120, inclusive. Therefore, using the sum of an arithmetic progression,
\(S=\frac{20}{2}(101+120)=10\times221=2210\). Hence, option B is correct. The value 2190 is the sum from 100 to 119, so it does not match the given range. Exam tip: Count terms in an inclusive range using \(\text{last term}-\text{first term}+1\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Since \(5050=\frac{100\times101}{2}\), we get \(n=100\). Therefore, \(S_{n-10}=S_{90}=\frac{90\times91}{2}=4095\). Option 4186 is not correct because it is the sum of the first 91 natural numbers. Exam tip: Recognise 5050 as \(S_{100}\), then reduce the subscript by 10.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{45}=\frac{45\times46}{2}=1035\) and \(S_{30}=\frac{30\times31}{2}=465\). Therefore, \(S_{45}:S_{30}=1035:465=69:31\). The option \(45:30\) is only the ratio of the numbers of terms, not the ratio of their sums. Exam tip: For ratios of such sums, apply \(\frac{n(n+1)}{2}\) first and then cancel common factors.

Here, \(S_n\) is the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), the first \(n-2\) terms cancel, leaving \((n-1)+n=2n-1\). Thus, \(2n-1=101\) gives \(n=51\). Therefore, \(S_{51}=\frac{51\times52}{2}=1326\). Option 1275 is \(S_{50}\), so it is close but not correct. Exam tip: In differences of partial sums, cancel the common initial terms and write the remaining terms.

There are \(45-15+1=31\) terms from 15 to 45. The sum of the first and last terms is \(15+45=60\), so the sum is \(\frac{31\times 60}{2}=930\). Although 960 is close, it does not follow from the correct number of terms and average of this sequence. Exam tip: When both endpoints are included, always use \(+1\) while counting the terms.

\(S_{n+1}=1+2+\cdots+n+(n+1)\) and \(S_{n-1}=1+2+\cdots+(n-1)\). On subtracting, the common terms cancel, leaving \(n+(n+1)=2n+1\). Hence, the correct answer is \(2n+1\). \(n+1\) is only the newly added last term; the term \(n\) must also be included. Exam tip: Expand the two sums and identify the terms that cancel.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{72}=\frac{72\times73}{2}=2628\) and \(S_{48}=\frac{48\times49}{2}=1176\). Hence, the difference is \(2628-1176=1452\). It is also the sum of the numbers from 49 to 72; an option such as 1462 results from an arithmetic error. Exam tip: for such questions, write \(S_{72}-S_{48}\) directly.