Sum of first n natural numbers

Using \(S_n=\frac{n(n+1)}{2}\), the sum is odd when n leaves remainder 1 or 2 on division by 4. Remainders 0 and 3 give an even sum. Exam tip: memorise this mod-4 pattern.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=1081\), we get \(n=46\). Hence, \(S_{n+4}-S_n\) is the sum of the next four terms: \(47+48+49+50=194\). Although 198 is close, it is not the correct sum of these four consecutive numbers. Exam tip: Rewrite \(S_{n+k}-S_n\) as the sum of terms from \(n+1\) to \(n+k\).

There are \(80-40+1=41\) terms from 40 to 80. Therefore, the sum is \(\frac{41(40+80)}{2}=41\times 60=2460\). Option 2420 may result from an error in counting the terms or finding the average. Exam tip: the number of integers from a first value to a last value is always \(\text{last}-\text{first}+1\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Here, \(S_n-S_{50}=1575\) and \(S_{50}=\frac{50\times51}{2}=1275\). Therefore, \(S_n=1575+1275=2850\). Now \(\frac{n(n+1)}{2}=2850\), so \(n(n+1)=5700=75\times76\). Hence, \(n=75\). The nearby option 80 does not give a difference of 1575 from \(S_{50}\). Exam tip: In such questions, first add the known \(S_k\) to find \(S_n\).

The sum of the first n natural numbers is n(n+1)/2. One of n and n+1 is even; after dividing by 2, the remaining factor must also be even. Thus the sum is even when n ≡ 0 or 3 (mod 4). Exam tip: check n modulo 4 directly.

If \(T=\frac{n(n+1)}{2}\), then \(8T+1=4n(n+1)+1=(2n+1)^2\), which is an odd perfect square. The converse is also true. \(4T+1\) has no such guaranteed form. Exam tip: test \(8T+1\).

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_{n+2}-S_n\), the common initial terms cancel, leaving \((n+1)+(n+2)=2n+3\). Thus, \(2n+3=155\), so \(2n=152\) and \(n=76\). If \(n=77\), the difference would be \(157\), not \(155\). Exam tip: In differences of sums, cancel the common terms and add only the remaining terms.

The sum of the first 120 natural numbers is \(\frac{120\times121}{2}=7260\), and the sum of the first 80 natural numbers is \(\frac{80\times81}{2}=3240\). Therefore, the required result is \(7260-3240=4020\). This is also the sum of the numbers from 81 to 120. An option such as 4040 can result from an arithmetic error in subtraction. Exam tip: use \(S_n=\frac{n(n+1)}{2}\) and subtract the two sums carefully.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{90}=\frac{90\times91}{2}=4095\) and \(S_{60}=\frac{60\times61}{2}=1830\). Therefore, \(4095:1830=273:122\). The ratio \(3:2\) is only the ratio of 90 to 60, not of their sums. Exam tip: When comparing such sums, include the \(n+1\) factor in \(n(n+1)/2\).

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=3655\), we get \(n=85\), since \(\frac{85\times86}{2}=3655\). Therefore, \(S_n-S_{n-4}=S_{85}-S_{81}\), which is the sum of the last four terms: \(82+83+84+85=334\). The value 330 would result from adding \(81+82+83+84\), which is not the required set of final four terms. Exam tip: \(S_n-S_{n-r}\) equals the sum of the last r terms up to n.

The sum is \(S=\frac{n(n+1)}{2}\). Hence \(8S+1=4n(n+1)+1=(2n+1)^2\), which is an odd perfect square. Exam tip: use this identity to test whether a number is triangular.

There are \(75-25+1=51\) terms from 25 to 75. These numbers form an arithmetic progression, so their sum is \(\frac{51}{2}(25+75)=\frac{51}{2}\times100=2550\). The option 2500 results from not using the number of terms or the average correctly. Exam tip: for an inclusive range from \(a\) to \(b\), always use \(b-a+1\) terms.

To obtain the sum of the first \(n\) natural numbers from \(S_{n-1}\), add the new term \(n\). Hence \(S_n=S_{n-1}+n\). Option B adds the previous term instead. Exam tip: compare consecutive partial sums to identify the added term.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{84}=\frac{84\times85}{2}=3570\). Half of this sum is \(\frac{3570}{2}=1785\), so option B is correct. A nearby value such as \(1815\) may look plausible, but it is not half of the required sum. Exam tip: first find the total using \(n(n+1)/2\), then apply the fraction asked for—in this case, one-half.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=4560\), we get \(n=95\). Hence, \(S_{n+5}-S_n\) is the sum of the next five natural numbers: \(96+97+98+99+100=490\). The value \(495\) would result if 101 were also included. Exam tip: \(S_{n+k}-S_n\) always represents the sum of terms from \(n+1\) to \(n+k\).

There are 30 terms from 121 to 150. This is an arithmetic progression, so
\(S=\frac{n}{2}(a+l)=\frac{30}{2}(121+150)=15\times271=4065\). Therefore, the correct answer is 4065. An answer such as 4035 may result from an error in counting the terms or using the last term. Exam tip: for a sum from \(a\) to \(l\), always count the terms as \(l-a+1\).

Adding the two equations gives (2S_n=1601), so (S_n=800.5). If an integer is not obtained, check the consistency of given values.

Let \(S_n\) be the sum of the first \(n\) natural numbers. In \(S_{n+1}-S_{n-2}\), the common terms from 1 to \(n-2\) cancel, leaving \((n-1)+n+(n+1)\). Hence, \(S_{n+1}-S_{n-2}=3n=150\), so \(n=50\). If \(n=49\), the difference would be 147, not 150. Exam tip: In differences of sums, cancel the common initial terms and write the remaining consecutive terms.

The sum of the first 100 natural numbers contains exactly one additional term beyond the sum of the first 99 natural numbers: 100. Therefore, \(S_{100}-S_{99}=100\). Option 99 is the last number among the first 99 natural numbers, not the difference between the two sums. Exam tip: The difference between sums of consecutive natural-number terms is always the next natural number.

Since \(S_n=\frac{n(n+1)}{2}\), we get \(\frac{S_n}{n}=\frac{n+1}{2}\). This is an integer only when \(n+1\) is even, so \(n\) must be odd. Being prime is not required; \(9\) also works. Exam tip: divide the sum formula by the given divisor first.

The sum of the first 150 natural numbers is \(\frac{150\times151}{2}=11325\), and the sum of the first 100 natural numbers is \(\frac{100\times101}{2}=5050\). Therefore, the difference is \(11325-5050=6275\). Equivalently, it is the sum of the numbers from 101 to 150. A nearby value such as 6175 results from an incorrect calculation. Exam tip: when subtracting two consecutive sums, the remaining terms begin at 101, not 100.

First identify n using Sₙ = n(n + 1)/2. We need n(n + 1)/2 = 1953, so n(n + 1) = 3906. Since 62 × 63 = 3906, n = 62. The requested index is therefore n + 2 = 64. Applying the sum formula gives S₆₄ = 64 × 65/2 = 32 × 65 = 2080. Hence option B is correct. Option A may result from adding only one new term incorrectly, while options C and D usually arise from advancing too far or making an arithmetic error. The essential step is to recover the original index before evaluating the shifted partial sum; the symbol n is not itself equal to 1953.

(S_n-S_{n-1}=n), so (n=84) and (S_{83}=\frac{83\cdot84}{2}=3486). For the previous sum, take (n-1).

Direct answer: option A, 399. First find the sum of the first 56 natural numbers: S₅₆ = 56 × 57/2 = 28 × 57 = 1596. The phrase 25% means 25/100, which simplifies to 1/4. Therefore, 25% of 1596 is 1596 ÷ 4 = 399. Option A is correct, and a check is 399 × 4 = 1596. Option B, 409, is not one-fourth of 1596 and may come from an addition or division error. Option C, 419, is similarly inconsistent with the percentage calculation. Option D, 429, is also too large for one-fourth of 1596. The two essential steps are first calculating the sum and then taking one-fourth of it. Memory cue: 25% always means one quarter, so divide the required quantity by 4.

Using \(S_n=\frac{n(n+1)}{2}\), we get \(8S_n+1=4n(n+1)+1=(2n+1)^2\), the square of an odd number. Option B is not a square for every \(n\). Exam tip: remember the \(8S_n+1\) property of triangular numbers.