Sum of first n natural numbers

Using the usual school convention, the natural numbers are 1, 2, 3, and so on. The condition x < 4 selects only the natural numbers that are strictly less than 4, namely 1, 2, and 3. The number 4 is excluded because the inequality is strict. Hence A = {1, 2, 3}.

Using (S_n=\frac{n(n+1)}{2}), (S_{10}=\frac{10\cdot11}{2}=55). In exams, put the correct value of (n).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{20}=\frac{20\times21}{2}=10\times21=210\). The value \(200\) may result from using \(20\times10\), but the correct formula must include \(n+1=21\). Exam tip: Natural numbers start from \(1\), so remember to use both \(n\) and \(n+1\) in the formula.

Using \(S_n=\frac{n(n+1)}{2}\) and substituting \(n=15\), \(S_{15}=\frac{15(15+1)}{2}=\frac{15\times16}{2}=15\times8=120\). Therefore, 120 is correct. An option such as 115 is incorrect because the formula includes the next number, \(16\), along with 15. Exam tip: write \(n+1\) first and simplify by dividing by 2 before multiplying.

The sum of the first (25) natural numbers is (\frac{25\cdot26}{2}=325). The formula saves time in long addition.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=30\), we get \(\frac{30\times31}{2}=15\times31=465\). Therefore, 465 is correct. The value 450 may result from an incorrect calculation rather than using the sum from 1 to 30. Exam tip: In such questions, remember to use both \(n\) and \(n+1\) in \(n(n+1)/2\).

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Substituting \(n=12\), \(S_{12}=\frac{12\times13}{2}=6\times13=78\). Therefore, 78 is the correct answer. A value such as 72 may result from using an incorrect multiplication or average. Exam tip: in such questions, use \(n(n+1)/2\) and divide the even factor by 2 before multiplying.

The sum of the first (n) natural numbers is (\frac{n(n+1)}{2}). Remembering this formula is the main key for this chapter.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=50\), so \(\frac{50\times51}{2}=25\times51=1275\). Therefore, the correct answer is 1275. The value 1250 results from using \(50\times25\) without including \(n+1=51\). Exam tip: For the sum from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The first 5 natural numbers are 1, 2, 3, 4, and 5. Therefore, \(1+2+3+4+5=15\), so option C is correct. Option 10 is the sum of only the first 4 natural numbers. Exam tip: The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

The first 8 natural numbers are 1 to 8. Their sum is found using \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_8=\frac{8(8+1)}{2}=\frac{8\times9}{2}=36\). The value 40 would result from adding 1 to 9 rather than 1 to 8. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=18\), so \(\frac{18\times19}{2}=9\times19=171\). Therefore, 171 is the correct answer. The value 180 may result from using \(18\times10\), but the correct formula uses \(n+1=19\). Exam tip: Write \(n(n+1)/2\) and divide the even factor by 2 first.

The governing concept is the sum formula for the first n natural numbers: S = n(n + 1)/2. Set S = 120, so n(n + 1)/2 = 120, which gives n(n + 1) = 240. Testing the positive integer options, n = 15 gives 15 × 16 = 240, and therefore S = 15 × 16 / 2 = 120. Hence option C is correct. The other choices do not satisfy the equation: n = 12 gives 78, n = 14 gives 105, and n = 16 gives 136. Since n counts natural numbers, a positive integer is required; the quadratic equation also confirms the valid root n = 15.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=40\), we get \(\frac{40\times41}{2}=20\times41=820\). Therefore, 820 is the correct answer. A value such as 800 may result from \(40\times20\), but that is not the correct sum formula. Exam tip: for the sum of consecutive natural numbers from 1 to \(n\), use \(\frac{n(n+1)}{2}\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=100\), so \(\frac{100\times101}{2}=50\times101=5050\). Therefore, 5050 is correct. The value 5000 results from using \(100\times50\) without the required adjustment for the \(101\) factor. Exam tip: For sums from \(1\) to \(n\), first write \(\frac{n(n+1)}{2}\), then substitute \(n\).

The first 7 natural numbers are \(1,2,3,4,5,6,7\). Their sum is \(1+2+3+4+5+6+7=28\). Using \(S_n=\frac{n(n+1)}{2}\), for \(n=7\), \(S_7=\frac{7\times8}{2}=28\). Note that 21 is the sum of the first 6 natural numbers. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=9\), we get \(\frac{9\times10}{2}=45\). Therefore, 45 is correct. The value 36 is the sum of natural numbers from 1 to 8, so it is a close but incorrect option. Exam tip: remember \(\frac{n(n+1)}{2}\) for the sum of the first \(n\) natural numbers.

Substitute \(n=6\) in the given formula: \(S_6=\frac{6(6+1)}{2}=\frac{6\times7}{2}=21\). Therefore, 21 is correct. A value such as 18 may result from substituting incorrectly. Exam tip: first find \(n+1\), then multiply and divide by 2.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Here, \(n=14\), so \(\frac{14\times15}{2}=7\times15=105\). Therefore, the correct answer is 105. The value 110 is not the sum of the numbers from 1 to 14. Exam tip: use \(n(n+1)/2\) and simplify the even factor with 2 first.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). For \(n=11\), \(\frac{11\times12}{2}=11\times6=66\). Therefore, the correct answer is 66. Note that 55 is the sum of the first 10 natural numbers. Exam tip: substitute the value of \(n\) carefully in \(n(n+1)/2\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(\frac{n(n+1)}{2}=55\), so \(n(n+1)=110\). Since \(10\times11=110\), \(n=10\). For instance, if \(n=9\), the sum is only \(\frac{9\times10}{2}=45\). Exam tip: use \(\frac{n(n+1)}{2}\) directly for sums of consecutive natural numbers.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{16}=\frac{16\times17}{2}=8\times17=136\). Hence, 136 is correct. A value such as 132 results from an incorrect calculation of \(16\times17\) divided by 2. Exam tip: divide the even factor by 2 first to calculate quickly.

The sum of the first \(n\) natural numbers is \(1+2+3+\cdots+n=\frac{n(n+1)}{2}\), so option A is correct. \(\frac{n(n-1)}{2}\) equals the sum of the first \(n-1\) natural numbers, making it a close but incorrect distractor. Exam tip: verify the formula with a small value such as \(n=3\); both the sum and \(\frac{3\times4}{2}\) give \(6\).

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=13\), we get \(\frac{13\times14}{2}=13\times7=91\). Therefore, 91 is correct. 84 is less than the actual sum of 13 natural numbers. Exam tip: For the sum from 1 to \(n\), directly use \(\frac{n(n+1)}{2}\).

The first 4 natural numbers are 1, 2, 3, and 4. Their sum is \(1+2+3+4=10\), so option C is correct. A value such as 12 may result from counting the numbers incorrectly. Exam tip: the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).