Sum of first n natural numbers

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).

The sum of the first 56 natural numbers is S₅₆ = 56 × 57/2 = 28 × 57 = 1596. Since 25% equals 25/100 = 1/4, the required value is 25% of 1596 = 1596/4 = 399. Therefore option A is correct. A useful check is that 400 × 4 = 1600, so one quarter of 1596 must be 399. The other choices are close values but do not equal one-fourth of the calculated sum; they may result from an incorrect sum formula, a percentage-conversion error, or faulty division. Both stages—finding S₅₆ and converting 25% to one-fourth—are essential.

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.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{64}=\frac{64\times65}{2}=32\times65=2080\). Hence, 2080 is correct. The value 2048 may result from using \(64\times32\), but that incorrectly omits the \((n+1)=65\) factor. Exam tip: Write \(n(n+1)/2\) and cancel the even factor with 2 first.

(S_{63}=\frac{63\times64}{2}=2016), so (n=63). Match (2S_n) with the product of two consecutive numbers.

The sum is (S_{72}-S_{30}=2628-465=2163). To find a middle range sum, subtract the previous terms.

By definition, S₈₅ = 1 + 2 + ... + 80 + 81 + 82 + 83 + 84 + 85, whereas S₈₀ = 1 + 2 + ... + 80. Subtracting cancels the common first 80 terms, leaving 81 + 82 + 83 + 84 + 85. Their sum is (81 + 85) × 5/2 = 166 × 2.5 = 415, or directly 81 + 82 + 83 + 84 + 85 = 415. Therefore option C is correct. The result is not the sum from 80 through 85 because the 80th term is already included in both partial sums and cancels. The other options reflect omission or incorrect inclusion of one of the remaining terms.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=465\) gives \(n(n+1)=930\), so \(n=30\). Therefore, \(S_{n+5}=S_{35}=\frac{35\times36}{2}=630\). Options such as \(635\) result from not applying the sum formula correctly for all terms from 31 to 35. Exam tip: first find \(n\) from the given sum, then substitute the new index in the formula.