Sum of first n natural numbers

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{15}=\frac{15\times16}{2}=120\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, \(120:55=24:11\). Although \(11:5\) is close, it is not the simplified form of \(120:55\). Exam tip: always divide both terms of a ratio by their greatest common factor to simplify it.

Direct answer: Sₙ₊₄ = 325, so option A is correct. Use Sₙ = n(n + 1)/2. We need to identify n from 231. Since 21 × 22/2 = 21 × 11 = 231, the given condition is S₂₁ = 231, so n = 21. The requested index is n + 4 = 25, not a change in the sum itself. Therefore calculate S₂₅ = 25 × 26/2 = 25 × 13 = 325. A direct check gives the same result: add the next four terms after 21 to S₂₁, namely 22 + 23 + 24 + 25 = 94; then 231 + 94 = 325. Option B, 351, is S₂₆ and may result from using one extra term. C and D correspond to still larger indices. The key distinction is between an index shift and adding a number to a sum. Memory cue: Sₙ₊₄ means the sum through the fourth next term, so first find n and then use n + 4 in the formula.

The sum is (S_{55}-S_{45}=1540-1035=505). You can also use the average of (10) consecutive terms.

The options should be checked carefully because (S_n-S_{20}=250) does not match the listed values. No listed option gives exactly (250).

The sum of the first n natural numbers is
\(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{28}=\frac{28\times29}{2}=406\) and \(S_{14}=\frac{14\times15}{2}=105\). Therefore, the difference is \(406-105=301\). An answer such as 291 usually results from an arithmetic error while subtracting the sum up to 14. Exam tip: for a difference of sums, calculate both sums using the formula and subtract the smaller one from the larger one.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=1540\), so \(n(n+1)=3080\). Since \(55\times56=3080\), \(n=55\). For example, if \(n=54\), the sum is \(\frac{54\times55}{2}=1485\), not 1540. Exam tip: find two consecutive factors of \(2S\) in such questions.

There are 10 numbers from 71 to 80, inclusive. Using the sum formula for an arithmetic sequence:

\(\frac{10}{2}(71+80)=5\times151=755\).

Therefore, the correct answer is 755. A choice such as 745 may result from an error in counting the number of terms or finding the average. Exam tip: the number of terms from one endpoint to another is \(\text{last term}-\text{first term}+1\).

The sum of the first n natural numbers is \\(\frac{n(n+1)}{2}\\). Thus, the sum of the first 20 natural numbers is \\(\frac{20\times21}{2}=210\\), and the sum of the first 5 natural numbers is \\(\frac{5\times6}{2}=15\\). Therefore, the required total is \\(210+15=225\\). The value 220 does not equal the sum of the first 20 natural numbers and may result from an incorrect calculation. Exam tip: calculate each sum separately before adding them.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=2850\), we get \(n=75\). Therefore, \(S_{n-5}=S_{70}=\frac{70\times71}{2}=2485\). The value 2556 is \(S_{71}\), so it is not correct here. Exam tip: first find \(n\) from the given sum, then calculate the sum for the required index.

Let Sₙ denote 1 + 2 + 3 + ⋯ + n, the sum of the first n natural numbers. The governing property is Sₙ − Sₙ₋₁ = n, because subtracting the sum through n − 1 leaves only the last term n. Therefore the given difference 68 tells us that n = 68. We then calculate S₆₈ using Sₙ = n(n + 1)/2: S₆₈ = 68 × 69/2 = 34 × 69 = 2346. Thus option B is correct. Option A is smaller than the required sum and can arise from an arithmetic error; option C equals S₆₉, while option D does not follow the formula. The difference identifies the index, and the formula gives the requested total.

When the sum is extended to \(n+1\) terms, only the new number \(n+1\) is added to the first \(n\) terms. Hence, \(S_{n+1}-S_n=n+1\). Option B gives the previous last term, not the added term. Exam tip: identify the newly included term.

If \(S_n\) denotes the sum of the first \(n\) natural numbers, then \(S_{36}=1+2+\cdots+36\). Adding the next two consecutive terms, 37 and 38, gives \(1+2+\cdots+36+37+38=S_{38}\). \(S_{37}\) includes terms only up to 37, so it is not correct. Exam tip: each next term added to a partial sum increases its subscript by 1.

Using \(S_n=\frac{n(n+1)}{2}\), one of \(n\) and \(n+1\) is even. After division by 2, the product is odd only when \(n\) leaves remainder 1 or 2 on division by 4. Thus, merely calling \(n\) odd or even is insufficient. Exam tip: test residues 0, 1, 2 and 3 modulo 4.

When the sum of the first 85 natural numbers is subtracted from the sum of the first 90 natural numbers, only the numbers from 86 to 90 remain. Thus, the difference is 86 + 87 + 88 + 89 + 90 = 440. Therefore, option C is correct. The value 450 would result from incorrectly treating each of the five terms as 90. Exam tip: In differences of such sums, cancel the common initial terms and add only the remaining consecutive terms.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=2211\), so \(n(n+1)=4422\). Since \(66\times67=4422\), \(n=66\). If \(n=67\), the sum would be \(\frac{67\times68}{2}=2278\), not 2211. Exam tip: multiply the given sum by \(2\) and look for two consecutive factors.

The sum \(1+2+3+\cdots+n\) is given by \(\frac{n(n+1)}{2}\). The expression \(\frac{n(n+1)(2n+1)}{6}\) is for the sum of squares. Exam tip: identify the sequence before selecting a formula.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{30}=\frac{30\times31}{2}=465\) and \(S_{20}=\frac{20\times21}{2}=210\). Therefore, \(S_{30}:S_{20}=465:210=31:14\). In 31:15, the second term does not match the simplified ratio. Exam tip: always divide both terms of a ratio by their greatest common factor to obtain the simplest form.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3240\), we get \(n=80\). Hence, \(S_{n+1}-S_n=S_{81}-S_{80}=81\), because the next sum includes only the next natural number. \(80\) is a close distractor, but it is the last number included in \(S_{80}\); the added number is \(81\). Exam tip: the difference of consecutive partial sums is always the next term.

There are \(10\) numbers from 81 to 90. Their average is \(\frac{81+90}{2}=85.5\), so the sum is \(10\times 85.5=855\). Hence, option B is correct. A value such as \(845\) can result from an error in counting the terms or finding the average. Exam tip: for consecutive numbers, use \(\text{number of terms}\times\text{average of the first and last terms}\).

Direct answer: n = 73, so option C is correct. The sum of the first n natural numbers is Sₙ = n(n + 1)/2. Set this equal to 2701: n(n + 1)/2 = 2701, so n(n + 1) = 5402. For n = 73, the consecutive product is 73 × 74. Compute it as 73 × (70 + 4) = 5110 + 292 = 5402. Hence S₇₃ = 5402/2 = 2701. The neighboring choices can be rejected by direct calculation: S₇₂ = 72 × 73/2 = 36 × 73 = 2628, while S₇₄ = 74 × 75/2 = 37 × 75 = 2775. Also, S₇₁ = 71 × 72/2 = 2556. Since each next sum adds a positive integer, the values increase strictly, so only 73 can give 2701. Memory cue: double the target sum and search for consecutive factors n and n + 1.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=40\), the sum is \(\frac{40\times41}{2}=820\). One-fourth of this is \(\frac{820}{4}=205\), so option B is correct. A value such as 195 results from an error in finding the sum or dividing it. Exam tip: first calculate the total using \(\frac{n(n+1)}{2}\), then take the required fraction.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{75}=\frac{75\times76}{2}=2850\) and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, the difference is \(2850-1275=1575\). Although 1585 is close, it does not result from the correct subtraction. Exam tip: this difference can also be viewed as the sum of the numbers from \(51\) to \(75\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=4950\), so \(n(n+1)=9900\). Since \(99\times100=9900\), \(n=99\). For \(n=100\), the sum would be \(5050\), not 4950. Exam tip: Multiply the given sum by 2 and look for two consecutive numbers whose product equals the result.

Direct answer: option C, 955. The numbers 91 through 100 form ten consecutive terms. Using the sum formula Sₙ = n(n + 1)/2, the required partial sum is S₁₀₀ − S₉₀. Now S₁₀₀ = 100 × 101/2 = 5050 and S₉₀ = 90 × 91/2 = 4095. Therefore, the required sum is 5050 − 4095 = 955. A second check is useful: the first and last numbers have average (91 + 100)/2 = 95.5, and there are 10 numbers, so the sum is 10 × 95.5 = 955. Option A, 935, is too small and may come from a subtraction error. Option B, 945, also results from an arithmetic mistake. Option C is correct because it includes every number from 91 through 100 exactly once. Option D, 965, is too large and may result from adding an extra amount. Memory cue: for equally spaced numbers, sum = number of terms × average.

(3160=S_{79}), so (S_{79}-S_{74}=75+76+77+78+79=385), not (390). Always match options with the final calculation.